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Karnataka 2nd PUC Chemistry Question Bank Chapter 15 Polymers

2nd PUC Chemistry Polymers NCERT Textbook Questions

Question 1.
Explain the terms polymer and monomer.
Answer:
Polymers are high molecular mass of macromolecules composed of repeating structural units derived from monomers. Polymers have a high molecular mass (10³-10⁷u). In a polymer, various monomer units are joined by strong covalent bonds. Polymers can be natural as well as synthetic. Polythene, rubber and nylon 6,6 are examples of polymers.

Monomers are simple, reactive molecules that combine with each other in large numbers through covalent bonds to give rise to polymers. For example ethene, propene, styrene, vinyl chloride.

Question 2.
What are natural and synthetic polymers? Give two examples of each type.
Answer:
Natural polymers are polymers that are found in nature. They are formed by plants and animals. Examples include protein, cellulose, starch etc.

Synthetic polymers are polymers made by human beings Examples include plastic (polythene), synthetic fibers (nylon 6, 6), synthetic rubbers (Buna-S).

Question 3.
Distinguish between the terms homopolymer and copolymer and give an example of each.
Answer:

Question 4.
How do you explain the functionality of a monomer?
Answer:
The functionality of a monomer is the number of binding sites that is/are present in that monomer.
For example, the functionality of monomers such as ethene and propene is one and that of 1, 3-butadiene and adipic acid is two.

Question 5.
Define the term polymerisation.
Answer:
Polymerisation is the process of forming high molecular mass (10³-10⁷u) macromolecules, which consist of repeating structural units derived from monomers. In a polymer, various monomer units are joined by strong covalent bonds.

Question 6.
Is ( NH-CHR-CO )₈, a homopolymer or copolymer?
Answer:
(NH CHR-CO)₈ is a homopolymer because it is obtained from a single monomer unit, NH₂-CHR-COOH.

Question 7.
In which classes, the polymers are classified on the basis of molecular forces?
Answer:
On the basis of magnitude of intermolecular forces present in polymers, they are classified into the following groups:

• Elastomers
• Fibers
• Thermoplastic polymers
• Thermosetting polymers

Question 8.
How can you differentiate between addition and condensation polymerisation?
Answer:
Addition polymerisation is the process of repeated addition of monomers, possessing double or triple bonds to form polymers. For example, polythene is formed by addition polymerization of ethene

Condensation polymerization is the process Of formation of polymers by repeated condensation reactions between two different bi functional or tri functional monomers. A small molecule such as water or hydrochloric acid is eliminated in each condensation. For example nylon 6, 6 is formed by condensation polymerisation of hexamethylene diamine and adipic acid.

Question 9.
Explain the term copolymerisation and give two examples.
Answer:
Formation of polymers from two or more different monomeric units is called copolymerization. Multiple units of each monomer are present in a copolymer. The process of forming polymer BunaS from 1, 3-butadiene and styrene is an example of copolymerization.

Nylon 6,6 is also a copolymer formed by hexamethylene diamine and adipic acid.

Question 10.
Write the free radical mechanism for the polymerisation of ethene.
Answer:
Polymerization of ethene to polythene consists of heating or exposing to a light mixture of ethene with a small amount of benzoyl peroxide as the initiator. The reaction involved in this process is given below:

Question 11.
Define thermoplastics and thermosetting polymers with two examples of each.
Answer:
Thermoplastics polymers are linear (slightly branched) long chain polymers, which can be repeatedly softened and hardened on heating. Hence, they can be modified again and again. Examples include polyethene, polystyrene.

Thermosetting polymers are cross-linked or heavily branded polymers which get hardened during moulding process. These plastics cannot be softened again on heating. Examples of thermosetting plastics include bakelite, urea-formaldelyde resins.

Question 12.
Write the monomers used for getting the following polymers.
1. Polyvinyl chloride
2. Teflon
3. Bakelite
Answer:
1. Vinyl chloride (CH₂ = CHCI)
2. Tetrafluro ethylene (CF₂ = CF₂)
3. Formaldehyde (HCHO) and phenol (C₆H₅OH)

Question 13.
Write the name and structure of one of the common initiators used in free radical addition polymerisation.
Answer:
One common initiator used in tree radical addition polymerization is benzoyl peroxide. Its structure is given below.

Question 14.
How does the presence of double bonds in rubber molecules influence their structure and reactivity?
Answer:
Natural rubber is a linear cis-polyisoprene in which the double bonds are present C₂ and C₃ of the isoprene units.

Because of this cis — configuration, inter molecular interactions between the various standard of isoprene are quite weak. As a result, various strands in natural rubber are arranged randomly. Hence it shows elasticity.

Question 15.
Discuss the main purpose of vulcanisation of rubber.
Answer:
Natural rubber though useful has some problems associated with its use. These limitations are discussed below;

• Natural rubber is quite soft and sticky at room temperature. At elevated temperatures (>335K), if becomes even softer. At low temperature (<283K) it becomes brittle. Thus to maintain its elasticity, natural rubber is generally used in the temperature range of 283K-335K.
• It has the capacity to absorb large amounts of water.
• It has low tensile strength and low resistance to abrasion.
• It is soluble in non polar solvents
• It is easily attacked by oxidizing agents
  Vulcanisation of natural rubber is done to improve all these properties. In this process, a mixture of raw rubber with sulphur and appropriate additive is heated at a temperature range between 373K and 415K.

Question 16.
What are the monomeric repeating units of Nylon-6 and Nylon-6,6?
Answer:
The monomeric repeating unit of nylon 6 is [NH – (CH₂)₅ – CO], which is derived form caprolactum.

The monomeric repeating unit of nylon 6, 6 is [NH – (CH₂)₆ – NH – CO – (CH₂)₄ – CO], which is derived from hexamethylene diamine and adipic acid.

Question 17.
Write the names and structures of the monomers of the following polymers:
(i) Buna-S
(ii) Buna-N
(iii) Dacron
(iv) Neoprene
Answer:

Question 18.
Identify the monomer in the following polymeric structures.

Answer:
Monomers are:

Question 19.
How is dacron obtained from ethylene glycol and terephthalic acid ?
Answer:
The condensation polymerisation of ethylene glycol and terephthalic acid leads to the formation dacron.

Question 20.
What is a biodegradable polymer ? Give an example of a biodegradable aliphatic polyester.
Answer:
A polymer that can be decomposed by bacteria is called a biodegradable polymer. Poly – β- hydroxybutyrate – Co -β – hydroxyvalerate (PHBV) is a biodegradable aliphatic polyester.

2nd PUC Polymers Additional Questions

Question 1.
How are polymers classified on the basis of solution?
Answer:

• Linear polymers such as high density polythene (HDP), Polyvinyl chloride, nylons, polyester etc.
• Branched polymers such as low density polythene (LDP), amylopectin, glycogen etc.

Question 2.
What is the difference between Buna-S and Buna-N ?
Answer:
Both are copolymers basically Buna-N is a copolymer of
1,3- butadiene and acrylonitrile. Buna-S is a copolymer of
1,3- butadiene and styrene.

Question 3.
What are the monomeric units of Nylon 6 and Nylon 6, 6?
Answer:
Monomeric repeating unit of Nylon 6 :

which is derived from caprolactam monomeric repeating unit of Nylon 6,6 :

which is derived from 2 monomers, hexamethylenediamine and adipic acid and has the following structure.

Question 4.
What are the disadvantages of natural rubber which are compromised by vulcanisation ?
Answer:

• It becomes more soft and sticky at high temperatures and brittle at low temperatures.
• It. has large water absorption capacity and low tensile strength and low resistance to abrasion.
• Not resistant to action of organic solvents.
• Easily attacked by oxygen and other oxidising agents.

Question 5.
What does PMMA stand for?
Answer:
Poly(methyl methacrylate)

Question 6.
How are polymers classified on the basis of molecular forces?
Answer:

• Elastomers
• Fibres
• Thermoplastic
• Thermosetting.

Question 7.
What is copolymerisation (and give an example)?
Answer:
When 2 or more different monomers are allowed to polymerise together, product is called copolymer and process is called copolymerisation.
Ex: Buna-S: 1, 3 – butadiene and styrene.

Question 8.
What thermoplastic and thermosetting polymers ?
Answer:
Thermoplastics are linear polymers which can be repeatedly softened on heating and hardened on cooling and hence can.be repeatedly used.
ex: polythene, polypropene etc.

Thermosetting polymers are on the other hand permanently setting polymers. On heating, they harden and cant be softened again.
ex: bakelite, urea-formaldehyde resin.

Question 9.
What is functionality of a polymer?
Answer:
It is the number of bonding sites in a molecule.
Functionality of styrene is one, whereas it is for adipic acid.

Question 10.
Write free radical mechanism for polymerisation of ethene.
Answer:
Initiation:

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Karnataka 1st PUC Physics Question Bank Chapter 11 Thermal Properties of Matter

1st PUC Physics Thermal Properties of Matter Textbook Questions and Answers

Question 1.
The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celcius and Fahrenheit scales.
Answer:
T_c = T – 273.15
∴ T_c for neon = 24.57 – 273.15
= – 248.58°C
and T_c for carbon dioxide
= 216.55 – 273.15
= – 56.60°C
t_F = \(\frac{9}{5}\) t_c + 32
∴ t_F for neon = \(\frac{9}{5}\) × (-248.58) + 32
= – 415.44°C
t_F for carbon dioxide
= \(\frac{9}{5}\) × (-56.60) + 32
= – 69.88°F

Question 2.
Two absolute scales A and B have triple points of water defined to be 200A and 350B. What is the relation between T_A and T_B?
Answer:
Triple point of water
= 273.16 K = 200A
= 350B
∴ Size of 1K on scale A = \(\frac{273.16}{200}\)
and size of 1K on scale B = \(\frac{273.16}{350}\)
Value of temperature T_A on scale A = \(\frac{273.16}{200}\) T_A
Value of temperature T_B on scale B = \(\frac{273.16}{350}\) T_B
Since the two temperatures are equal,
\(\left(\frac{273.16}{200}\right) \mathrm{T}_{\mathrm{A}}=\left(\frac{273.16}{350}\right) \mathrm{T}_{\mathrm{B}}\)
⇒ T_A = \(\frac{4}{7}\) T_B

Question 3.
The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law: R = R₀ [1 + α (T – T₀)] The resistance is 101.6Ω at the triple point of water 273.16K, and 165.5 Ω at the normal melting point of lead (600.5K). What Is the temperature when the resistance is 123.4 Ω?
Answer:
At T₀ = 273.16K, R₀= 101.6 Ω
At T₁ = 600.5K, R₁ = 165.5 Ω
Let T₂ be the temperature at which the resistance is R₂ = 123.4 Ω
R = R₀ [1 + α (T – T₀)]
⇒ T – T₀ = \(\frac{\mathrm{R}-\mathrm{R}_{0}}{\alpha \mathrm{R}_{0}}\)
∴ T₁ – T₀ = \(\frac{\mathrm{R}_{1}-\mathrm{R}_{0}}{\alpha \mathrm{R}_{0}}\)→ (1)
T₂ – T₀ = \(\frac{\mathrm{R}_{2}-\mathrm{R}_{0}}{\alpha \mathrm{R}_{0}}\)→ (2)
Dividing (2) by (1) gives,
\(\frac{\mathrm{T}_{2}-\mathrm{T}_{0}}{\mathrm{T}_{1}-\mathrm{T}_{0}}=\frac{\mathrm{R}_{2}-\mathrm{R}_{0}}{\mathrm{R}_{1}-\mathrm{R}_{0}}\)
\(\mathrm{T}_{2}=\mathrm{T}_{0}+\left(\mathrm{T}_{1}-\mathrm{T}_{0}\right)\left(\frac{\mathrm{R}_{2}-\mathrm{R}_{0}}{\mathrm{R}_{1}-\mathrm{R}_{0}}\right)\)
= 293.16 + (600.5-273.16) \(\left(\frac{123.4-101.6}{165.5-101.6}\right)\)
⇒ T₂ = 384.83 K

Question 4.
Answer the following:
1. The triple point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the
boiling point of water as standard fixed points (as was originally done in the Celcius scale)?

2. There were two fixed points in the original Celcius scale as mentioned above which were assigned the number 0°C and 100°C respectively. On the absolute scale, one of the fixed points is the triple point of water, which on the Kelvin absolute scale is assigned the number 273.16K. What is the other fixed point on this (Kelvin) scale?

3. The absolute temperature (Kelvin scale) T is related to the temperature te on the Celcius scale by t_c = T – 273.15 Why do we have 273.15 in this relation and not 273.16?

4. What is the temperature of the triple point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale?
Answer:
1. The triple point is used as a standard because of the unique conditions of temperature and pressure at this point. The melting point of ice and boiling point of water depend on pressure. So they cannot be used as reference points.

2. The other fixed point is absolute zero, the theoretically lowest possible temperature.

3. We have 273.15 in the relation and not 273.16 because the triple point of water is 0.01 °C and not 0°C.
In other words, 0.01 °C → 273.16 K
⇒ 0°C → 273.15 K
∴ t_c = T – 273.15

4. The unit interval size in the Fahrenheit scale = \(\frac{212-32}{100}\) =1.8
∴ The triple point of water = 273.16 × 1.8
= 491.69

Question 5.
Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made:

1. What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B?
2. What do you think Is the reason behind the slight difference in the answers of thermometers A and B? (The thermometers are not faulty) What further procedure is needed In the experiment to reduce the discrepancy between the two readings?
Answer:
1. Let P be the pressure in the pressure thermometer and T be the corresponding absolute temperature.
Let P₀ and T₀ be the values for the triple point of water.
Then \(\frac{T}{T_{0}}=\frac{P}{P_{0}}\)
\(\therefore \frac{T_{A}}{273.16}=\frac{1.797 \times 10^{5}}{1.250 \times 10^{5}}\)
⇒ T_A = 392.69 K
\(\frac{T_{B}}{273.16}=\frac{0.287 \times 10^{5}}{0.200 \times 10^{5}}\)
⇒ T_B= 391.98 K

2. The discrepancy is due to the non¬ideal behaviour of the gases. To reduce this discrepancy, measurements at lower pressures should be taken.

Question 6.
A steel tape 1m long Is correctly calibrated for a temperature of 27.0°C. The length of a steel rod measured by this tape Is found to be 63.0 cm on a hot day when the temperature is 45.0°C. What is the actual length of the steel rod on that day? What is the length of the same steel rod on a day when the temperature is 27.0° C? Coefficient of linear expansion of steel = 1.20 × 10^-5K^-1.
Answer:
l₀ = 63 cm
∆T = 45°C – 27°C .
= 18°C = 18 K
α = 1.2 × 10^-5K^-1
l = l₀ (1 + α ∆T)
= 63(1 + 1.2 × 10^-5 × 18)
l = 63.0136 cm.
∴ Actual length of the steel rod at 45°C =63.0136 cm
At 27°C, the steel tape is correctly calibrated.
∴ the length of the steel rod = 63.0 cm

Question 7.
A large steel wheel Is to be fitted on to a shaft of the same material. At 27°C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole In the wheel is 8.69 cm. The shaft Is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft? Assume the coefficient of linear expansion of the steel to be constant over the required temperature range: α_steel = 1.2 × 10^-5K^-1.
Answer:
Given α_steel = 1.2 × 10^-5K^-1
l₁ = 8.70 cm
l₂ = 8.69 cm (because the shaft is cooled to help it slip on the wheel)
T₁ = 27°C = 300 K
T₂ =?
l₂ = l₁ [1 + α_steel(T₂ – T₁)]
⇒ 8.69 = 8.70 [1 + 1.2 × 10^-5(T₂ – 300)]
⇒ 1 + 1.2 × 10^-5(T₂ – 300) = \(\frac{8.69}{8.70}-1\)
⇒ T₂ = 300 + \(\frac{-0.01}{8.70} \times \frac{1}{1.2 \times 10^{-5}}\)
= 204.21 K
= – 68.94°C
= – 69°C
∴ The required temperature = – 69°C

Question 8.
A hole drilled in a copper sheet. The diameter of the hole is 4.24cm at 27.0°C. What is the change in the diameter of the hole when the sheet is heated to 227° C? Coefficient of linear expansion of copper = 1.70 × 10^-5K^-1.
Answer:
Given α = 1.70 × 10^-5K^-1
β = 2α = 3.40 × 10^-5K^-1
d₁ = 4.24 cm
Initial area A₁ at 27°C = 300K = T₁
is \(\pi\left(\frac{\mathrm{d}_{1}}{2}\right)^{2}=\pi\left(\frac{4.24}{2}\right)^{2}\)
⇒ A₁ = 4.4944 π cm²
Area A₂ at 227°C = 500K = T₂ is
∆₁ [1 + β(T₂ – T₁)]
i.e. A₂ = 4.4944 π [1 + 3.40 × 10^-5 (500 – 300)]
A₂ = 4.5250 π cm²
⇒ \(\pi\left(\frac{\mathrm{d}_{2}}{2}\right)^{2}\) = 4.5250 π cm²
⇒ d₂ = \(\sqrt{18.100}\) =4.2544 cm
∴ Change in diameter = d₂ – d₁
= 4.2544 – 4.25 = 0.0144 cm

Question 9.
A brass wire 1.8 m long at 27°C Is held taut with little tension between two rigid supports. If the wire Is cooled to a temperature of -39°C, what Is the tension developed In the wire, if its diameter is 2.0 mm?
Coefficient of linear expansion of brass = 2.0 × 10^-5 K^-1, Young’s modulus Of brass = 0.91 × 10¹¹ Pa.
Answer:
l = 1.8 m,
T₁ =27°C = 300 K,
T₂ = – 39°C =234 K,
d = 2mm = 2 × 10^-3
α_brass = 2.0 × 10^-5 K^-1,
Y_brass = 0.91 × 10¹¹ Pa
\(\frac{\Delta \ell}{\ell}=\alpha \Delta \mathrm{t}\)
\(Y=\frac{\Delta F / A}{\Delta \ell / \ell}=Y\left(\frac{\Delta \ell}{\ell}\right) A\)
= Y(α ∆T) \(\pi\left(\frac{\mathrm{d}}{2}\right)^{2}\)
= 0.91 × 10¹¹ × 2.0 × 10^-5 α (234 – 300) × π × \(\left(\frac{2 \times 10^{-3}}{2}\right)^{2}\)
= – 377.37 N
Tension developed ∆F = 380N (positive value, rounded to 2 significant figures)

Question 10.
A brass rod of length 50cm and diameter 3.0 mm is Joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250°C, if the original lengths are at 40.0°C? Is there a ‘thermal stress’ developed at the junction? The ends of the rod are free to expand (Coefficient of linear expansion of brass = 2.0 × 10^-5K^-1, steel = 1.2 × 10^-5K^-1).
Answer:
Given:  l_brass = 50cm, l_steel = 50cm
T₁ = 40°C = 313K
T₂ = 250°C = 523 K
α_brass = 2.0 × 10^-5K^-1
α_steel = 1.2 × 10^-5K^-1
∆l_brass = l_brass × α_brass × ∆T
= 50 × 2.0 × 10^-5 × (523 – 313)
∆l_brass = 0.21 cm
∆l_steel = l_steel × α_steel × ∆T
= 50 × 1.2 × 10^-5 × (523 – 313)
= 0.126 cm
∆l_steel = 0.13cm
(round off to two significant figures) Since the rods are free to expand, no thermal stress is developed at the junction.

Question 11.
The coefficient of volume expansion of glycerin is 49 × 10^-5K^-1. What is the fractional change in its density for a 30°C rise In temperature?
Answer:
Given, r_glycerin = 49 × 10^-5K^-1
∆T = 30°C = 30 K
Let the volume be V₁ at temperature and V₂ at temperature T₂
Using V₂ = V₁(1 + r ∆T), we get
\(\frac{V_{1}}{V_{2}}=\frac{1}{1+r \Delta T}\)
\(=\frac{1}{1+49 \times 10^{-5} \times 30}=0.9855\)
Fractional change in density

= 0.9855 – 1
= – 0.0145
= – 1.5 × 10^-2
(The negative sign denotes a decrease in density).

Question 12.
A10 kW drilling machine is used to drill a bore In a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the blocks in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings? Specific heat of aluminium = 0.91 J g^-1 K^-1
Answer:
Given,
P = 10kW,
t = 2.5 minutes = 150s,
m = 8 kg,
s_aluminium = 0.91 Jg^-1 K^-1 = 910 Jkg^-1 K^-1
Energy transferred to the aluminium
= 50% of p × t
This energy heats up the aluminium.
So, \(\frac{50}{100}\) × pt = m s_aluminum ∆T
⇒ \(\frac{1}{2}\) × 10 × 10³ × 150 = 8 × 910 × ∆T
⇒ ∆T = 103 K
or ∆T = 103°C

Question 13.
A copper block of mass 2.5kg is heated in a furnace to a temperature of 500_°C and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper = 0.39 Jg^-1 K^-1 heat of fusion of water = 335 Jg^-1)
Answer:
Given m_copper = 2.5 kg,
T₁= 500°C,
T₂ = 0°C,
S_copper = 0-39 Jg^-1 K^-1
L_ice = 335 Jg^-1
Let the mass of ice that melts be m_ice
Heat dissipated by the copper block = Heat absorbed by the ice.
⇒ m_copper S_copper ∆T = m_ice L_ice
⇒ 2500 × 0.39 × (500 – 0) = m_ice × 335
⇒ m_ice = 1455 g
⇒ m_ice = 1.5 kg
(rounding off to 2 significant figures)

Question 14.
In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150°C Is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm³ of water at 27°C. The final temperature is 40°C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value of the specific heat of the metal?
Answer:
Given, mass of metal block,
m₁ = 0.20 kg
Initial temperature of the metal block,
T₁ = 150°C
Initial temperature of the calorimeter and water,
T₂ = 27_°C
Final temperature of the mixture,
T₃ = 40°C
Volume of water in the calorimeter,
V =150 cm³
= 150 × 10^-6 m³
Mass of the water = Density × Volume
= 1000 × 150 × 10^-6
= 150 × 10^-3 kg
Let s₁ be the specific heat of the metal.
s₂ is the specific heat of water.
Water equivalent of the calorimeter,
W = 0.025 kg
Heat lost by the metal block = Heat gained by the calorimeter and water.
⇒ m₁ s₁ ∆T₁ = (m₂ + W) s₂ ∆T₂
⇒ 0.2 × s₁ × (150 – 40)
= (150 × 10^-3 + 0.025) × 4200 × (40 – 27)
⇒ s₁ = 434.3 J kg^-1K^-1
= 430 J kg^-1K^-1
or 0.43 Jg^-1K^-1
If the heat losses to the surroundings are not negligible, the left-hand side of the expression would have had an extra term added to it to account for this loss. Upon solving for s₂ we would get a smaller value than the actual value.

Question 15.
Given below are observations on molar specific heats at room temperature of some common gases.
Gas                         Molar specific heat (cv) (cal mol’1 K’1)
Hydrogen               4.87
Nitrogen                 4.97
Oxygen                   5.02
Nitric oxide             4.99
Carbon monoxide   5.01
Chlorine                  6.17
The measured molar specific heats of these gases are markedly different from those for monoatomic gases. Typically, the molar specific heat of a monoatomic gas is 2.92 cal/mol K. Explain this difference. What can you Infer from the somewhat larger (than the rest) value for chlorine?
Answer:
All the gases mentioned above are diatomic, and hence have other degrees of freedom in addition to the translational degrees of freedom. To raise the temperature of the gas, heat has to be supplied to increase the average energy of all the modes.

Consequently, the molar specific heats for diatomic gases is higher than that for monoatomic gases. The higher value of the molar specific heat of chlorine indicates that at room temperature, vibrational modes are also present In addition to the rotational modes of freedom.

Question 16.
Answer the following questions based on the P-T diagram of carbon dioxide:
1. At what temperature and pressure can the solid, liquid and vapour phases of Co₂, exist in equilibrium?
2. what is the effect of a decrease of pressure on the fusion and boiling point of Co₂?
3. What are the critical temperature and pressure for Co₂? What is their significance?
4. Is Co₂ solid, liquid or gas at

• – 70°C under 1 atm,
• – 60°C under 10 atm,
• 15°C under 56 atm?

Answer:

1. At -56.6°C and 5.11 atm, which is the triple point of Co₂, all 3 .phases can coexist in equilibrium.
2. From the fusion and vaporisation curves, we.see that reducing the pressure lowers the fusion and boiling point.
3. The critical temperature and pressure are 31.1°C and 73.0 atm respectively. Above this temperature, Co₂ will not liquefy even if compressed to high pressures.
4.

• Gas
• Solid
• Liquid

Question 17.
Answer the following questions based on the P-T phase diagram of Co₂:

1. Co₂ at 1 atm pressure and temperature – 60°C is compressed isothermally. Does it go through a liquid phase?
2. What happens when Co₂ at 4 atm pressure is cooled from room temperature at constant pressure?
3. Describe qualitatively the changes In a given mass of solid Co₂ at 10 atm pressure and temperature -65°C as it is heated up to room temperature at constant pressure.
4. Co₂ is heated to a temperature 700C and compressed isothermally. What changes in its properties do you expect to observe?

Answer:

1. When subjected to isothermal compression, Co₂ will condense to solid directly without passing through the liquid phase.
2. When cooled at constant pressure, Co₂ will condense directly to solid phase without passing through the liquid phase,
3. As the solid Co₂ is heated up, it first turns to liquid phase and then solid phase. The fusion and boiling points can be obtained from finding out the points of intersection between the horizontal 10 atm line on the P-T diagram with the fusion and vaporisation curves.
4. Co₂ will not exhibit any clear transition to the liquid phase but will deviate more and more from ideal behaviour ps the pressure increases.

Question 18.
A child running a temperature of 101°F is given an antipyrin (i.e., a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98°F in 20 min, what is the average rate of extra evaporation caused by the drug? Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of the human body is approximately the same as that of water, and latent heat of evaporation of water ‘at that temperature is about 580 cal g^-1.
Answer:
Mass of the child, m = 30 kg
Change in temperature ∆T ,
= 101°F – 98°F
= 3°F
= 3 × 5/9 = 5/3°c
Specific heat of water,
s = 4.2 × 103 J kg^-1°C^-1
Latent heat of vaporisation,
L = 580 cal g^-1
= 580 × 4.2 × 10³ J kg^-1
Time taken for the sweat to evaporate,
t = 20 min.
Let m₁ be the mass of sweat that evaporates.
Now, heat lost by the child during evaporation
= Heat needed for the sweat to evaporate.
⇒ ms ∆T = m₁L
⇒ 30 × 4.2 × 10³ × 5/3
= m₁ × 580 × 4.2 × 10³
⇒ m₁ = 0.0862 kg
m₁ = 86.2 g
Rate of extra evaporation caused by the drug = Rate of evaporation of sweat
\(=\frac{m_{1}}{t}=\frac{86.2 g}{20 m i n}\)
= 4.3 g min^-1

Question 19.
A ‘thermacole’ icebox is a cheap and efficient method for storing small quantities of cooked food In summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg pt Ice is put in the box, estimate the amount of Ice remaining after 6h. The outside temperature is 45°C, and co-efficient of thermal conductivity of thermacole is 0.01 J s^-1 m^-1K^-1. [Heat of fusion of water = 835 × 10³ J kg^-1]
Answer:
Total surface area of the cube,
A = 6 × (30 cm)²
= 0.54 m²
Mass of ice in the box M = 4 kg
Thickness of the box, L = 5 cm = 0.05 m
Thermal conductivity of thermacole,
k = 0.01 Js^-1m^-1K^-1.
Outside temperature T₁ = 45°C
Temperature of ice T₂ = 0°C
Let the mass of ice that melts be m kg.
Heat of fusion of water,
L_{f ice} = 335 × 10³ J kg^-1
Time t = 6h = 6 × 3600 s = 21600 s
Heat gained by ice during fusion = Heat conducted through the walls of the box
⇒ m L_{f, ice} = \(\frac{\mathrm{KA}\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right) \mathrm{t}}{\mathrm{L}}\)
⇒ m × 335 × 10³
\(=\frac{0.01 \times 0.54 \times(45-0)}{0.05} \times 21600\)
⇒ m = 0.31 kg
∴ Mass of ice left in the box = M – m
= 4 – 0.31
= 3.69
= 3.7 kg.

Question 20.
A brass boiler has a base area of 0.15 m² and thickness 1.0 cm. It bolls water at the rate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 J s^-1m^-1K^-1; Heat of vaporisation of water = 2256 × 10³ J kg^-1.
Answer:
Let the temperature of the flame be T₁
T₂ = boiling point of water = 100°C
Thermal conductivity of brass,
k = 109 J s^-1m^-1K^-1
Base area of the boiler, A = 0.15 m²
Thickness of the boiler, L = 1 cm = 0.01 m
⇒ Heat of vaporisation of water
L_steam = 2256 × 103 J kg^-1
Rate of boiling water
= 6 kg min^-1 = \(\frac{6}{60}\) kg s^-1
= 0.1 kg s^-1
∴ Power consumed = 0.1 × L_steam
= 0.1 × 2256 × 10³W
= 2256 × 10²W
∴ Powe transmitted through the boiler
= 2256 × 10² W
\(\Rightarrow \frac{\mathrm{KA}\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)}{\mathrm{L}}\) = 2256 × 10²
⇒ \(\frac{109 \times 0.15 \times\left(T_{1}-100\right)}{0.01}\) = 2256 × 10²
⇒ T₁ – 100 = 138
⇒ T₁ = 238°C
∴ Temperature of the flame = 238°C

Question 21.
Explain why:
1. a body with large reflectivity Is a poor emitter.

2. a brass tumbler feels much colder than a wooden tray on a chilly day.

3. an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open but gives a correct value for the temperature when the same piece is in the furnace.

4. The earth without its atmosphere would be inhospitably cold.

5. heating systems based on circulation of steam are more efficient in warming a building than those based on the circulation of hot water.
Answer:
1. We know that a + r + 1 = 1, where ‘a’ is absorbance, ‘r’ is reflectance and ‘t’ is transmittance or emittance. According to Kirchhoff’s law, emittance ∝ absorbance, i.e., good absorbers are also good emitters and so are poor reflectors. Therefore, if reflectivity is large (‘r’ is large) then ‘a’ is small and hence emittance is smaller (the object behaves as a poor emitter).

2. The coefficient of thermal conductivity of brass is higher than that of wood. When a brass tumbler is touched, heat quickly flows from human body to the tumbler and so the tumbler appears colder. But in the case of the wooden tray heat does not flow from the human body to the wooden tray, and so it feels relatively hotter.

3. Let the temperature of the hot iron in the furnace be‘T’ and that of the external environment be ‘T₀’. According to Stefan’s Law, heat energy radiated per second E = σT⁴ in the open. Clearly E > E’. Hence the low readings of the optical pyrometer when the iron piece is in the open.

4. The earth’s atmosphere acts like an insulating blanket around it and does not allow heat to escape out but reflects it back to the earth. If it were absent, the heat would have escaped and the planet would be very cold.

5. Steam has a higher heat capacity (due to the latent heat) than boiling water. Therefore steam-based water heating systems are more effective.

Question 22.
A body cools from 80°C to 50°C in 5 minutes. Calculate the time it takes to cool from 60°C to 30°C. The temperature of the surroundings is 20°C.
Answer:
The average temperature of 80° C and 50° C is 65° C, which is 45° C above the room temperature. Under these conditions, the body cools 30°C in 5 minutes. Using the equation.
\(\frac{\text { Change in temperature }}{\text { Time }}=\mathrm{KT}\)
we have \(\frac{30^{\circ} \mathrm{C}}{5 \mathrm{min}}\) = K (45° C) ……(1)
The average of 60° C and 30° C is 45° C which is 25° C above the room temperature. K is the same for this situation as the original.
Hence \(\frac{30^{\circ} \mathrm{C}}{\mathrm{t}}\) = K (25° C) ……(2)
Divide (1) by (2), we get,

∴ Time taken = 9 minutes.

1st PUC Physics Thermal Properties of Matter One Mark Questions and Answers

Question 1.
Define coefficient of thermal conductivity.
Answer:
Coefficient of thermal conductivity of a material is defined as the quantity of heat conducted per second through a unit area of a slab of unit thickness when the temperature difference between its ends is 1K.

Question 2.
What is thermal radiation?
Answer:
Heat radiation from a body due to its temperature is called thermal radiation.

Question 3.
What is a perfect black body?
Answer:
A body, which absorbs all the radiation incident on it is called a perfect black body.

Question 4.
Define emissivity of a surface.
Answer:
Emissivity of a surface is defined as the ratio of the quantity of thermal radiation emitted per sec per unit area of the surface to the quantity of thermal radiation emitted per sec per unit area of a perfect black body under identical conditions.

Question 5.
Define absorptivity of a surface.
Answer:
Absorptivity of a surface is defined as the ratio of the quantity of thermal radiation absorbed per sec to the quantity of radiation incident on it per second.

Question 6.
What is the unit of coefficient of thermal conductivity?
Answer:
Wm^-1K^-1

Question 7.
What is the dimensional formula of coefficient of thermal conductivity?
Answer:
MLT^-3 θ^-1

Question 8.
What happens to the heat supplied to a good conductor when it is under steady-state?
Answer:
Heat is entirely conducted from the hot end to the cold end.

Question 9.
Why is coal tar heated before use?
Answer:
Coal tar is heated to reduce its viscosity.

Question 10.
What is the value of emissivity of a perfect black body?
Answer:
Emissivity of a perfect black body is unity.

Question 11.
Give dimensions of solar constant and water equivalent.
Answer:
Solar constant: [MT^-3]
Water equivalent: [M]

1st PUC Physics Thermal Properties of Matter Two Marks Questions and Answers

Question 1.
Explain three methods of transfer of heat.
Answer:

1. Conduction:
It is a process of transmission of heat in solids, without the actual movement of the particles.

2. Convection:
It is a process of transmission of heat in fluids (liquids and gases) with the actual movement of particles.

3. Radiation:
It is a process of transmission of heat without any medium.

Question 2.
Explain steady-state and temperature gradient.
Answer:
When a solid is heated at one end, heat starts flowing from hot end to cold end. Heat supplied to any element of the solid is partly used in heating it, partly conducted through it and rest is radiated from its surface. After a certain stage, there is no variation of temperature with time and heat supplied is entirely conducted and radiated.

Such a state is called steady state. Under steady-state, the temperature at different parts of the solid remains constant and is independent of time. The ratio of the difference in temperature to the distance between two points in a heated body under steady state is called the temperature gradient.

Question 3.
Mention the properties of thermal radiation.
Answer:
Properties of thermal radiation:

1. It travels through a vacuum with the velocity of light.
2. It travels in a straight line.
3. It undergoes reflection, refraction and total internal reflection.
4. It exhibits the phenomenon of interference, diffraction and polarisation.
5. Intensity of radiation decreases with increase in distance.

Question 4.
Explain Ferry’s black body.
Answer:

Ferry’s black body consists of a hollow double-walled sphere with a fine hole and a pointed projection in front of the hole. The space between the two walls is evacuated to reduce any loss of radiation due to convection. The inner walls are coated with lamp black. Any radiation falling on the hole is completely absorbed due to multiple reflections, hence the hole acts as a black body.

Question 5.
State and explain Kirchhoff’s law.
Answer:
At a given temperature, the ratio of the emissive power to the absorptive power for all bodies is a constant and is equal to the emissive power of a perfect black body.
If e is the emissive power and a is the absorptive power then, \(\frac{\mathrm{e}}{\mathrm{a}}\) = constant  =1.
∴ e = a(∵ emissive power of a perfectly black body = 1)
Thus, the emissive power of a body is equal to its absorptive power, which means good emitter of thermal radiation is also a good absorber of heat.

Question 6.
State and explain Newton’s law of cooling.
Answer:
The rate of cooling of a body is directly proportional to difference in temperature of the body and the surrounding. If the temperature of a body
decreases from θ₁ to θ₂ in t seconds, then rate of cooling is, \(\frac{\theta_{1}-\theta_{2}}{t}\)
Average temperature of the body θ = \(\frac{\theta_{1}+\theta_{2}}{2}\)
According to Newton’s law,
\(\frac{\theta_{1}-\theta_{2}}{t} \propto\left(\theta-\theta_{0}\right)\)
θ₀ – temperature of the surrounding.

Question 7.
State and explain Stefan’s law.
Answer:
The total radiation emitted by a perfect black body per second per unit area of the surface is directly proportional to the fourth power of its absolute temperature. If E is radiation emitted per second per unit area of a perfect black body at temperature T then, E ∝ T⁴ or E = σT⁴ where s is called Stefan’s constant.

Question 8.
State and explain Wein’s displacement law.
Answer:
The wavelength (λ_m) corresponding to maximum energy emitted by a black body is inversely proportional to its absolute temperature.
i. e., λ_m ∝ \(\frac{1}{T}\) or λ_m T = constant.
The constant is called Wein’s Constant.

Question 9.
State and explain Planck’s law.
Answer:
According to this law, radiation of a particular frequency (v) is emitted or absorbed in discrete amounts (E) given by the relation E = h v where h is a Planck’s constant. The value of h is 6.625 × 10^-34 Js. These packets of energy are called quantum or photon.

Question 10.
Define solar constant. What is its value?
Answer:
The solar constant is defined as the amount of heat radiation absorbed normally per second per unit area of a perfect black body placed on the earth at a mean distance from the sun in the absence of atmosphere.
The value of solar Constant is 1350 W/m².

Question 11.
State

1. Kirchhoff’s law of radiation,
2. State Stefan’s law of radiation

Answer:
1. Kirchhoff’s law:
At a given temperature, the ratio of the emissive power to the absorptive power is a constant for all bodies and is equal to the emissive power of a perfect black body.

2. Stefan’s law:
The total radiation emitted by a perfectly black body per second per unit area is directly proportional to the 4^th power of its absolute temperature.

Question 12.
Calculate the heat of combustion of coal, when 10 gm of coal, on burning raises the temperature of 2 litres of water from 20° C to 55° C.
Answer:
mass of 2l of water = 2kg
∆T = 55° C – 20° C = 35° C
s = 4.2 × 10³ J kg^-1 °C^-1
Heat produced by burning coal = ms ∆T
= 2 × 4.2 × 10³ × 35
= 2.94 × 10⁵ J

Question 13.
2 kg of water at 80° C is mixed with 3 kg of water at 20°C. Assuming no heat losses, find the final temperature of the mixture.
Answer:
m₁ = 2 kg,
T₁ = 80°C,
s = 4.2 × 10³ J kg^-1 °C^-1,
m₂ = 3kg,
T₂ = 20° C
Let the final temperature be T₃
Heat lost by 3 kg water = Heat gained by 2 kg water.
⇒ m₁s(T₁ – T₃) = m₂s(T₃ – T₂)
⇒ 2 × (80 – T₃) = 3 × (T₃ – 20)
⇒ T₃ = 44° C
∴ Final temperature = 44° C

Question 14.
What are the basic requirements of a cooking utensil in respect of:

1. Specific heat and
2. thermal conductivity

Answer:
For the temperature of the utensil to rise quickly, it is necessary for the utensil to have low specific heat. In order for the heat to spread quickly to the vegetables, the utensil must have high thermal conductivity.

1st PUC Physics Thermal Properties of Matter Three Marks Questions and Answers

Question 1.
State Wien’s displacement law. Draw the graph showing energy emitted versus wavelength for a black body at different temperatures.
Answer:
Wein’s displacement law:
The wavelength (λ_m) Corresponding to maximum energy emitted by a black body is inversely proportional to its absolute temperature. i.e. λ_m ∝ \(\frac{1}{T}\) or λ_m T = b. where b is a constant, called Wein’s constant. The radiation emitted by a black body is called black body radiation. The distribution of energy of a black body radiation at different temperatures is as shown in the figure.

The energy distribution is not uniform. There is a particular wavelength λ_m at which the energy emitted is maximum. The wavelength λ_m for which the intensity is maximum decreases with increase in temperature.

Question 2.
Draw energy distribution curves for a black body at two different temperatures T₁ and T₂ (T₁ > T₂). Write any two conclusions that can be drawn from these curves.
Answer:
The radiation emitted by a black body is called black body radiation. The distribution of energy of a black body radiation at different temperatures with its wavelength is as shown in the figure.

Question 3.
Draw labelled curves which show how the energy in the black body radiation spectrum varies with wavelength at various temperatures. What result does this lead to?
Answer:
The radiation emitted by a black body is called black body radiation. The distribution of energy of a black body radiation at different temperatures with its wavelength is as shown in the figure.

1st PUC Physics Thermal Properties of Matter Four Or Five Marks Questions and Answers

Question 1.
State and explain the laws of thermal conductivity.
Answer:

Consider a metal slab of area of cross-section A and thickness d. Let the end faces (ABCD & EFGH) be maintained at temperatures θ₁ and θ₂ (θ₁ > θ₂).
The amount of heat conducted (Q) from the hotter to colder face is,
1. directly proportional to the area of cross-section (A)
2. directly proportional to the time for which heat flows (t)
3. directly proportional to the temperature difference between the faces (θ₁ – θ₂)
4. inversely proportional to the distance between the faces (d)
This is called the law of thermal conductivity.
According to the law Q ∝ \(\frac{\mathrm{At}\left(\theta_{1}-\theta_{2}\right)}{\mathrm{d}}\)
\(\mathrm{Q}=\frac{\mathrm{KAt}\left(\theta_{1}-\theta_{2}\right)}{\mathrm{d}}\)
K is a constant of proportionality called coefficient of thermal conductivity.

Question 2.
What is black body radiation? Explain the characteristics of It.
Answer:
The radiation emitted by a black body is called black body radiation. The distribution of energy of a black body radiation at different temperatures with its wavelength is as shown in the figure.

Characteristics of black body radiation

1. A black body emits radiation of all wavelengths lying in the region of ultraviolet, visible light and infrared.
2. As the temperature of the black body increases, the total amount of radiation emitted by it increases. (According to Stefan’s law)
3. The energy distribution is not uniform. There is a particular wavelength λ_m at which the energy emitted is maximum.
4. The wavelength λ_m for which the energy emitted is maximum decreases with increase in temperature.

Question 3.
Describe a total radiation pyrometer.
Answer:

It consists of a concave mirror M made of copper and plated with nickel. The mirror has a small hole at the centre at which an eyepiece E is fitted. A diaphragm D is placed at the focus of M; which has a hole to allow radiation to pass through it. A blackened metal strip S is placed behind D. To the other surface of S, one junction of a thermocouple TT is connected. The thermocouple is connected to a millivoltmeter. A screen R is used to protect the strip from direct radiation. The mirror can be adjusted for focusing.

Question 4.
Why is ice packed in sawdust? Explain.
Answer:
Ice is packed in sawdust because it prevents ice from melting. Sawdust is a bad conductor of heat.

Question 5.
Explain the different modes of transmission of heat. Explain Wein’s displacement law with the help of graph of intensity versus wavelength.
Answer:
1. Conduction:
It is a process of transmission of heat in solids, without the actual movement of the particles.
Eg: When metal rod is heated at one end, after some time another end also becomes hot.

2. Convection:
It is a process of transmission of heat in fluids (liquids and gases) with the actual movement of particles.
Eg: Smoke carries heat energy from the flame by convection.

3. Radiation:
It is a process of transmission of heat without any medium in the form an electromagnetic waves.
Eg: Heat from the sun reaches the earth by radiation.

Wein’s displacement law:
The wavelength (λ_m) Corresponding to maximum energy emitted by a black body is inversely proportional to its absolute temperature. i.e. λ_m ∝ \(\frac{1}{T}\) or λ_m T = b. where b is a constant, called Wein’s constant. The radiation emitted by a black body is called black body radiation. The distribution of energy of a black body radiation at different temperatures is as shown in the figure.

The energy distribution is not uniform. There is a particular wavelength λ_m at which the energy emitted is maximum. The wavelength λ_m for which the intensity is maximum decreases with increase in temperature.

Question 6.
What is a black body? Draw the curves showing the energy distribution among black body radiations at different temperature. Hence define Wein’s displacement law. Give one application of Wein’s displacement law.
Answer:
1. black body:
The radiation emitted by a black body is called black body radiation. The distribution of energy of a black body radiation at different temperatures with its wavelength.

Wein’s displacement law:
The wavelength (λ_m) Corresponding to maximum energy emitted by a black body is inversely proportional to its absolute temperature. i.e. λ_m ∝ \(\frac{1}{T}\) 17 or λ_m T = b. where b is a constant, called Wein’s constant.

Question 7.
Discuss briefly energy distribution of black body radiation. Hence deduce Wien’s displacement law and Stefan’s law.
Answer:
Wein’s displacement law:
The wavelength (λ_m) Corresponding to maximum energy emitted by a black body is inversely proportional to its absolute temperature. i.e. λ_m ∝ \(\frac{1}{T}\) or λ_m T = b. where b is a constant, called Wein’s constant. The radiation emitted by a black body is called black body radiation. The distribution of energy of a black body radiation at different temperatures is as shown in the figure.

The energy distribution is not uniform. There is a particular wavelength λ_m at which the energy emitted is maximum. The wavelength λ_m for which the intensity is maximum decreases with increase in temperature.

Stefan’s law:
The total radiation emitted by a perfect black body per second per unit area of the surface is directly proportional to the fourth power of its absolute temperature. If E is radiation emitted per second per unit area of a perfect black body at temperature T then, E ∝ T⁴ or E = σT⁴ where s is called Stefan’s constant.

1st PUC Physics Thermal Properties of Matter Numerical Problems Questions and Answers

Question 1.
A circular plate has a radius of 10 cm and thickness 1 cm. If one face of the plate is kept in contact with steam and the other face is in contact with ice, calculate the amount of heat flowing per hour across the plate. K as 85 W^-1m^-1K^-1.
Solution:
The amount of heat conducted in t seconds is given by,
\(Q=K \frac{A\left(\theta_{1}-\theta_{2}\right) t}{d}\)
Here, K = 85 W m^-1K^-1
A = π r² = 3.14 × (0.1)²
t = 1 hr = 60 × 60 s.
d = 10^-2 m²
(θ₁ – θ₂) = 373 -273 = 100 K
Q= \(\mathrm{K} \frac{\mathrm{A}\left(\theta_{1}-\theta_{2}\right) \mathrm{t}}{\mathrm{d}}\)
= \(\frac{85 \times 3.14 \times(0.1)^{2} \times 100 \times 60 \times 60}{10^{-2}}\)
= 96.08×10⁶ J.

Question 2.
A metal rod 0.314 m long and 0.02 m in diameter has one end in boiling water at 373K and the other end in melting ice at 273K. Find the amount of ice that will melt per hour? K = 385 Wm^-1K^-1.
Latent heat of Ice = 3.36 × 10⁵ Jkg^-1.
Solution:
If ‘m’ is mass of ice that melts in one hour, then quantity of heat conducted is Q = mL
But Q = K \(\frac{A\left(\theta_{1}-\theta_{2}\right) t}{d}\)
∴ Q = mL
m = \(\frac{\mathrm{KA}\left(\theta_{1}-\theta_{2}\right) t}{\mathrm{d} \times L}\)
Here, K =385Wm^-1K^-1
A = p r² =3.14 × (0.01 )²
L = 3.3 × 10⁵ Jkg^-1
θ₁ – θ₂ = 373 – 273 = 100 K
t = 1 hour = 3600 s
d = 0.314 m
∴ m= \(\frac{385 \times 3.142 \times(0.01)^{2} \times 100 \times 60 \times 60}{0.314 \times 3.36 \times 10^{5}}\)
= 41.28 x 10^-2kg

Question 3.
Two metal rods of the same area of cross-section and with their lengths in the ratio 1:3 are joined together to form a compound rod. One end of this compound rod is kept at 100°C and the other at 0° C. Calculate the temperature at the Junction if the thermal conductivity is in the ratio 1:3.
Solution:

Let AB and BC be the two rods of length L₁ and L₂ and having the same area of cross-section A, joined together to form a compound rod as shown in the figure.
If ‘ θ ’ is the steady temperature at the junction B during the steady-state, then, rate of flow of heat through AB = rate of flow through BC

Question 4.
Calculate rate of heat radiated per unit area from the surface of a furnace at a temperature 2000K, assuming it to be a black body. If the emissivity is 0.37, then what Is the amount of heat radiated per second? G =5.67×10^-8Wm^-2K^-4.
Solution:
From Stefan’s law of radiation, the amount of heat radiated per second is E = σT⁴.
E =5.67 × 10^-8 × (2000)⁴
= 5.67 × 2⁴ × 10⁴
= 5.67 × 16 × 10⁴
= 9.072 × 10⁵Js^-1.
If e is the emissivity of the surface, then the heat radiated per second R¹ = eR
= 0.37 × 9.072 × 10⁵
= 3.35 × 10⁵ Js^-1

Question 5.
A slab of stone of area 0.36m² and thickness 0.1m is exposed on the lower surface to steam at 373 K. A block of ice at 273 k rests on the upper surface of the slab. If in one hour, 4.8kg of Ice is melted, calculate the conductivity of the stone. Latent heat of Ice = 3.35 x 10⁵ JK^-1.
Solution:
A = 0.36m²,
d = 0.1m
θ₁ = 373 K
θ₂ = 273 K
t = 1 hr = 3600s,
m = 4.8kg,
K = ?
L = 3.35 × 10⁵ Jkg^-1
Heat conducted through the rod = Heal used in melting ice
\(\frac{\mathrm{KA}\left(\theta_{1}-\theta_{2}\right) \mathrm{t}}{\mathrm{d}}\) = mL
K = \(\frac{m L d}{A\left(\theta_{1}-\theta_{2}\right) t}=\frac{4.8 \times 3.35 \times 10^{5} \times 0.1}{0.36 \times 100 \times 3600}\)
= 1.241 Wm^-1 K^-1.

Question 6.
The filament of 40 W electric lamp of length 0.1m and diameter 0.1mm is kept at a temperature of 2773 K. Calculate the emissive power if all the heat is lost by radiation. σ = 5.67 × 10^-8 Wm^-2 K^-4].
Solution :
P = 40 W,
l = 0.1 m,
r = 0.05 mm
T = 2773 K
e = ?
Area = 2π rl =3.142 x 10^-5 m²
Energy emitted by a source in t sec
E = e A σ T⁴t
\(\frac{E}{t}\) = e A σ T⁴
P = e A σ T⁴
e = \(\frac{P}{A \sigma T^{4}}\)
\(=\frac{40}{3.142 \times 10^{-5} \times 5.67 \times 10^{-8} \times(2773)^{4}}\)
= 0.3798

Question 7.
The thickness of ice in a lake at a certain moment is 0.03m. At what rate the thickness of the ice increasing and how long will it take for the thickness of ice layer to be doubled, if the temperature of air is -20° C. Given coefficient of thermal conductivity of ice=0.168 Wm^-1K^-1 degree. Latent heat of fusion of ice=3.35 × 10 Jkg^-1 density of ice= 9.2 × 10² Kg/m³
Solution:
When the layer of ice covers the lake, the lower part of the ice is in contact with ice is at 0°c and outer surface with air at – 20°c
∴ θ₁ =0°c and θ₂ = – 20°c
Let the thickness of ice increase by small amount ∆x by an area A in a time ∆t.
Then,
mass of ice formed = A × ∆x × ρ
Quantity of heat, Q = A × ∆x × ρ × L    …..(1)
But, Q = \(\frac{\mathrm{k} \mathrm{A}\left(\theta_{1}-\theta_{2}\right) \Delta \mathrm{t}}{\mathrm{d}}\) ….(2)
from equation (1) & equation (2)
A × ∆x × r × L = \(\frac{\mathrm{k} \mathrm{A}\left(\theta_{1}-\theta_{2}\right) \Delta \mathrm{t}}{\mathrm{d}}\)
\(\frac{\Delta x}{\Delta t}=\frac{k A\left(\theta_{1}-\theta_{2}\right)}{d x A x \rho x L}\)
\(=\frac{0.168 \times 20}{0.03 \times 9.2 \times 10^{2} \times 3.35 \times 10^{5}}\)
= 3.634 × 10^-7ms^-1
The time taken for the ice layer to double is
t = \(\frac{\Delta x / \Delta t}{d}\)
\(=\frac{3.634 \times 10^{-7}}{0.06}\)
= 60.57 × 10^-7sec

Question 8.
Estimate the rate at which ice melts in a wooden box 0.04m thick, of inside measurement 0.6 m × 0.4m × 0.4m. Assume that the outside temperature is 300 K and the conductivity of wood is 0.08 wm^-1K^-1
Solution:
Area of the box
= A = 4 × (0.6 × 0.4) + 2 × (0.4 × 0.4)= 1.28m²
d = 0.04m
k= 0.08wm^-1K^-1
L= 3.36 × 10⁵Jkg^-1
Let Q be the heat transmitted per second.
Then, Q = \(\frac{\mathrm{k} \mathrm{A}\left(\theta_{1}-\theta_{2}\right) \mathrm{t}}{\mathrm{d}}\)
\(=\frac{0.08 \times 1.28 \times(300-273) \times 1}{0.04}=69.12 J\)
To melt 1 kg of ice, 3.36 × 10⁵J of heat is required
∴ Rate at which ice melts
\(=\frac{69.12}{3.36 \times 10^{-5}}\)
= 20.57 × 10^-5kgs^-1.

Question 9.
A man wraps himself in flannel 4 mm thick. How much heat will be lost per square metre of his body per hour if the atmosphere temperature is 20° and the body temperature is 37°c(The co-efficient of thermal conductivity of flannel is 0.05 Wm^-1k^-1)
Solution:
k=0.05Wm^-1k^-1
d = 4mm = 4 × 10³
m = 0.004m
θ₁ =37°c,
θ₂ =20°c
t =1 hr=3600 seconds
A =1m²
The quantity of heat conducted is
\(Q=\frac{k A\left(\theta_{1}-\theta_{2}\right) t}{d}\)
\(=\frac{0.05 \times 1(37-20) 3600}{0.004}\)
= 7.65 × 10⁵J

Question 10.
Calculate the rate at which heat is lost through a glass window of area 0.1m² thickness 4mm when the temperature inside is at 40° c and outside is -5°c. Co-efficient of thermal conductivity of a glass is 1.1wm^-1k^-1.
Solution:
A=0.1m²
d = 4mm = 4 × 10^-3
m=0.004m
θ₁ = 40°C
θ₂ = -5°c
k = 1.1Wm^-1k^-1
Heat lost through the glass window is
\(Q=\frac{K A\left(\theta_{1}-\theta_{2}\right) t}{d}\)
\(=\frac{1.1 \times 0.1(40+5) \times 1}{0.004}\)
= 1237.5J

Question 11.
The two faces of metal plate of thickness 40mm and area 10²m² maintained at 273K and 373K. The coefficient of thermal conductivity of the metal is 380 Wm^-1k^-1 Find the quantity of heat flows, through the plate in one hour.
Solution:
d = 40mm = 40 × 10^-3
m = 0.04m
A = 10²m²
θ₂ = 273k
θ₁ =37K
K = 380Wm^-1k^-1
t = 1hr = 3600 seconds
The quantity of heat that flows in one hour is
\(Q=\frac{k A\left(\theta_{1}-\theta_{2}\right) t}{d}\)
\(=\frac{380 \times 10^{2}(373-273) \times 3600}{0.04}\)
Q = 3.42 × 10¹¹J

Question 12.
A spherical body of radius 0.01m is at 400K. Calculate the energy radiated by the body per second per unit area. Assume the body to be perfect black body.
Solution:
From stefan’s law of radiation, amount of heat radiated per second per unit area is
E = σ T⁴
= 5.67 × 10^-8 × (400)⁴ = 1451.5s^-1
The total energy radiated per second is
E = A σ T⁴
E = P r² σ T⁴
= 3.142 × (0.01 )² × 5.67 × 10^-8 × (400)⁴
= 0.4561s^-1

Question 13.
A metal rod 0.3m long & 0.02m in diameter has one end at 100°C & other end at 0°C. Calculate the total amount of heat conducted in 2 minutes. Given k = 385 JS^-1m^-1K^-1.
Solution:
l = 0.3m,
d = 0.02m
∴ r = 0.01m
θ₁ = 100°C
θ₂ = O°C
t = 2 minutes = 120 S
k = 385 JS^-1m^-1K^-1.
Q = ?
t= π r² = 3.142 × 0.01² = 3.142 × 10⁴m²
\(\therefore Q=\frac{k A t\left(\theta_{1}-\theta_{2}\right)}{\ell}\)
\(=\frac{385 \times 3.142 \times 10^{4} \times 120(109)}{0.3}\)
= 4.8 × 10³ J

Question 14.
A metallic rod 1.5m long and 2×10⁴ m² in area is heated at one end. The coefficient of thermal conductivity of the material of the metallic rod is 210 Js^-1m^-1K^-1. In the steady-state the temperature of the ends of the rod are 373K and 263 K. Calculate.

1. Temperature gradient In the rod.
2. Temperature at a point 0.8 m from the hot end.
3. Rate of heat transmission.

Answer:
d=1.5m,
A = 2 × 10⁴m²,
K = 210 W/m/k, θ₁ = 373K and θ₂ = 263K.
1. Temperature gradient
\(\frac{\theta_{1}-\theta_{2}}{d}=\frac{373-263}{1.5}=\frac{110}{1.5}=73.33 \mathrm{K} / \mathrm{m}\)

2. Let θ be the temperature at a distance d¹ = 0.8m from the hot end then temperature gradient = \(\frac{\theta_{1}-\theta}{d^{1}}\)
i.e., 73.33 = \(\frac{373-\theta}{0.8}\) or θ =314.336 K.

3. Rate of heat transmission
\(\frac{\mathrm{q}}{\mathrm{t}}=\mathrm{KA}\left(\frac{\theta_{1}-\theta_{2}}{\mathrm{d}}\right)\)
= 210 × 2 × 10^-4 × 73.33
= 3.08 J/s.

Question 15.
The two faces of a metal plate of thickness 40 mm and area 10^-2m² are maintained at 273 K and 373 K. The coefficient of thermal conductivity of metal is 380 w/m/K. Find the quantity of heat that flows through the plate in one hour.
Answer:
d = 40 × 10^-3m,
A = 10^-2m²,
t = 1 hr = 3600s,
θ₁ = 373K,
θ₁ = 2.73K and K= 380 W/m/K
\(\mathrm{q}=\frac{\kappa \mathrm{At}\left(\mathrm{q}-\theta_{2}\right)}{\mathrm{d}}\)
= 34.6×10⁶J.

Question 16.
The time taken for a substance to cool form 100° C to 50° C is 5 minutes. If the room temperature is 30°C, what time required for same substance to cool from 50°C to 40°C?
Answer:
Case (i):
Rate of cooling
\(\frac{\mathrm{d} \theta}{\mathrm{dt}}=\frac{100-50}{5}\) = 10° C/min,
Average temperature
\(\theta=\frac{100+50}{2}=75^{\circ} \mathrm{C}\)
Room temperate θ₀ = 30°C.
According to Newton’s law of cooling,
\(\frac{\mathrm{d} \theta}{\mathrm{dt}}\) = -K(θ – θ₀).
10 = -K(75 – 30).
K= \(-\frac{10}{45} / \mathrm{min}\) . ……(1).

Case(ii):
Rate of cooling
\(\frac{\mathrm{d} \theta}{\mathrm{dt}}\) = \(\frac{50-40}{\mathrm{dt}}=\frac{10}{\mathrm{dt}}\),
Average temperature
\(\theta=\frac{50+40}{2}=45^{\circ} \mathrm{C}\)
Room temperate e θ₀ = 30°C.
According to Newton’s law of cooling,
\(\frac{\mathrm{d} \theta}{\mathrm{dt}}\) = K(θ – θ₀)
\(\frac{10}{\mathrm{dt}}=-\left(\frac{-10}{45}\right)(45-30)\) (from)
dt =3 min.

1st PUC Physics Thermal Properties of Matter Hard Questions and Answers

Question 1.
A hot body placed in air is cooled down. According to Newton’s law of cooling, the rate of decrease of temperature being ‘k’ times the temperature of the surrounding. Starting from t = 0, find the time in which the body will lose 90% of the maximum heat it can lose.
Answer:
We have,
\(\frac{\mathrm{d} \mathrm{T}}{\mathrm{dt}}\) = – K(T – T₀)
where T₀ is the temperature of the surrounding and T is the temperature of the body at time t. Let T = T₁ at t = 0.
Then

⇒ T – T₀ = (T₁ – T₀) e^-kt ……(1)
The body continues to lose heat until its temperature becomes equal to that of the surrounding. The loss of heat in this entire period is
∆H = ms(T -T₀).
This is the maximum heat it can lose. If the body loses 90% of this heat, the decrease in to temperature will be
\(\frac{9 \Delta \mathrm{H}}{10 \mathrm{ms}}\) = \(\frac{9}{10}\) (T₁ – T₀)
If the body loses this heat in time t, the temperature at t₁ will be
T₁ – \(\frac{9}{10}\) (T₁ – T₀)
\(=\frac{T_{1}}{10}+\frac{9 T_{0}}{10}\)
Putting these values of time and temperature in (1),
\(\frac{T_{1}+9 T_{0}}{10}\) – T₀ = (T₁ – T₂) e^-kt₁
⇒ \(\frac{T_{1}-T_{0}}{10}\) = (T₁ – T₂) e^-kt₁
⇒ e^-kt₁ = 1/10
⇒ t₁ = \(\frac{\ln 10}{k}\)
∴ Required time = \(\frac{\ln 10}{k}\)

Question 2.
Two thin metallic spherical shells of radii r₁ and r₂ (r₁ < r₂ are placed with their centres coinciding. A material of thermal conductivity k is filled in the space between the shells. The inner shell is maintained at temperature T₁ and the outer shell at temperature T₂ (T₁ < T₂). Calculate the rate at which heat flows radially through the material.
Answer:

Consider two spherical shells of radii x and x + dx concentric with the given system. Let the temperatures at these shells be T and T + dt respectively. The amount of heat flowing radially inward through the material between x and x + dx is

Karnataka 2nd PUC Kannada Workbook Answers Pallava Chapter 3 Nanarthagalu

Karnataka 2nd PUC Chemistry Question Bank Chapter 1 The Solid State

Question 1
Define the term ‘amorphous’. Give a few examples of amorphous solids.
Answer:
Amorphous solids are those solids in which the constituent particles may have short range order but do not have a long range order. They have irregular shapes and are isotropic in nature. They do not undergo a clean cleavage.They do not have sharp melting points or definite heat of fusion. E.g.: Glass, rubber and plastics.

Question 2.
What makes a glass different from a solid such as quartz? Under what conditions could quartz be converted into glass?
Answer:
Glass is an amorphous solid in which the constituent particles (SiO₄ tetrahedra) have only short range order and there is no long range order. On melting quartz and then cooling it rapidly, it is converted into glass.

Question 3.
Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous.
(i) Tetra phosphorus deioxide (P₄O₁₀)
(ii) Ammonium phosphate (NH₄)₃PO₄
(iii) SiC
(iv) I₂
(v) P₄
(vi) Plastic
(vii) Graphite
(viii) Brass
(ix) Rb
(x) LiBr
(xi) Si
Answer:
Ionic – (NH₄)₃PO₄ and LiBr
Metallic – Brass and Rb
Molecular – P₄O₁₀,I₂ P₄
Network (Covalent) – Graphite, SiC, Si
Amorphous – Plastics

Question 4.
(i) What is meant by the term ‘coordination number’?
(ii) What is the coordination number of atoms:
(a) in a cubic close-packed structure?
(b) in a body-centred cubic structure?
Answer:
(i) The coordination number of a constituent particles (atom, ion or molecule) in a crystal is the number of constituent particles which are the immediate neighbour of the particle in the crystal. In ionic crystals coordination number of an ion in the crystal is the number of oppositely charged ions surrounding that particular ion.
(ii) a] 12 b] 8.

Question 5.
How can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell? Explain.
Answer:
The atomic mass can be determined using the formula

where
ρ is the density of the metal, a is the dimension of the unit cell, N₀ is Avogadro number and Z is the number of atoms in the unit cell.

Question 6.
(a) ‘Stability of a crystal is reflected in the magnitude of its melting points’.
(b) The melting points of some compounds are given below:
Water = 273K, ethyl alcohol = 155.7 K, Diethyl ether = 156.8K, Methane = 90.5K.
What can you say about intermolecular forces between these molecules?
Answer:
(a) Higher the melting point, greater are tire “forces holding the constituent particles together and hence greater is the stability.

(b) The intermolecular forces in water and ethyl alcohol are mainly hydrogen bonding. Higher melting point of water than alcohol shows that hydrogen bonding in ethyl alcohol molecules is not as strong as in water molecules. Diethyl ether is a polar molecule. The intermolecular forces present in them are dipole -dipole attraction. Methane is a non-polar molecule. The only forces present in them are the weak Vander Waal’s forces (London / dispersion forces).

Question 7.
How will you distinguish between the following pairs of terms:
(i) Hexagonal close-packing and cubic close-packing?
(ii) Crystal lattice and unit cell?
(iii) Tetrahedral void and octahedral void?
Answer:
(i) When a third layer is placed over the second layer in such a way that the spheres cover the tetrahedral or ‘C’ voids, a three dimensional closest packing is obtained where the spheres in every third or alternate layers are vertically aligned (i.e. the third layer is directly above the first, the fourth above the second layer or so on) calling the first layer as layer A and the second layer as layer B, the arrangement is ABAB pattern or hexagonal close packing.

When the third layer is placed over the second layer in such a way that spheres cover the octahedral or ‘d’ voids, a layer differentfrom layers A and B is produced. Let us call it as layer C, continuing further, a packing is obtained were the spheres in every fourth layer will be vertically aligned. This pattern of stocking spheres is called ABCABC… pattern or close cube packing (CCP).

(ii) A regular three dimensional arrangement of points in space is called crystal lattice.
Unit cell is the smallest portion of a crystal lattice which, when repeated in different directions, generates the entire lattice.

(iii) Wherever a sphere of the second layer is above the void of the first layer (or vice versa) a tetrahedral void is formed. These voids are called tetrahedral voids because a tetrahedron is formed when the center of these four spheres arejoined.
The voids which are surrounded by six spheres is called octahedral voids.

Question 8.
How many lattice points are there in one unit cell of each of the following lattice?
(i) Face-centred cubic
(ii) Face-centred tetragonal
(iii) Body-centred
Ans:
Lattice points in face centred cubic or face centred tetragonal
= 8 (at corners) + 6 (at face centres) = 14
However, particles per unit cell
8 × \(\frac{1}{8}+\) + 6 × \(\frac{1}{2}+\) =4
lattice points in body centred cube = 8 (at comers) + 1 (at body centre) = 9.
However, particles per unit cell = 8 × \(\frac{1}{8}+\) + 1 = 2

Question 9.
Explain:
(i) The basis of similarities and differences between metallic and ionic crystals.
(ii) Ionic solids are hard and brittle.
Answer:
(a) Similarities (i) Both ionic and metallic crystals have electrostatic forces of attraction. In ionic crystals, these are between the oppositely charged ions. In metals, these are among valence electrons and the kernels. That is why both have high melting point.

(ii) in both cases the bond is non-directional Differences: (i) In ionic crystals, the ions are not free to move. Hence, they cannot conduct electricity in the solid state. They can do so only in the matter state or in aqueous solutions. In metals the electrons are free to flow. Hence they can conduct electricity in the solid state.

(ii) Ionic bond is strong due to electrostatic forces of attraction.
Metalic bond may be weak or strong depending on the number of valance electrons and size of the kernels.

Question 10.
Calculate the efficiency of packing in case of a metal crystal for
(i) simple cubic
(ii) body-centred cubic
(iii) face-centred cubic
(with the assumptions that atoms are touching each other).
Answer:
(i)

Top/Bottom/side view of simple cubic unit cell As spheres are touching each other, evidently a = 2 r
Number of spheres per unit cell =\(\frac { 1 }{ 2 }\) × 8 = 1
volume of the sphere = \(\frac { 4 }{ 3 }\)πr³
volume of the cube = a³ = (2r)³ = 8r³
∴ Fraction occupied, i.e. packing fraction
(\(\frac { 4 }{ 3 }\)πr³) / 8r³
or % occupied = 52.4%.

(ii)

spheres at the corners, body diagonal, AD = 4r further, face diagonal,

Number of spheres per unit cell = 8 × \(\frac { 1 }{ 8 }\) + 1 = 2
volume of 2 spheres = 2 × \(\frac { 4 }{ 3 }\) πr³ = \(\frac { 8 }{ 3 }\) πr³
∴ Fraction occupied i.e. packing fraction

(iii)

A,B,C are the centres of the sphere . As sphere on the face centre is touching the spheres at the corners evidently AC = 4r
But from right angled triangle ABC

∴ volume of the unit cell

Number of spheres in the unit cell
= 8 × \(\frac { 1 }{ 8 }\) + 6 × \(\frac { 1 }{ 2 }\) = 4
volume of 4 spheres = = 4 × \(\frac { 4 }{ 3 }\) πr³ = \(\frac { 16 }{ 3 }\) πr³
∴ fraction occupied, i.e. packing fraction

or % occupied = 74%.

Question 11.
Silver crystallises in fee lattice. If edge length of the cell is 4.07 x 10^-8 cm and density is 10.5 g cm^-3, calculate the atomic mass of silver.
Answer:

Question 12.
A cubic solid ¡s made of two elements P and Q. Atoms of Q are at the corners of the cube and P at the bodycentre. What is the formula of the compound? What are the coordination numbers of P and Q?
Answer:
As atoms Q are present at the 8 corners of the cube, therefore, number of atoms of in the unit cell = \(\frac { 1 }{ 8 }\) × 8 = 1
As atoms P are present at the body centre, therefore, number of atoms of x in the unit cell = 1
∴Ratio of atoms x:y =1:1
Hence the fonnula of the compound is PQ coordination number of each of P and Q = 8.

Question 13.
Niobium crystallises in body-centred cubic structure. If density is 8.55 g cm^-3, calculate atomic radius of niobium using its atomic mass 93 u.

or x = antilog0.519 = 3.304 for body – centred cubic,
r = \(\frac{\sqrt{3}}{4}\) a = 0.433a = 0.433 x 330.4 pm = 143.1 pm

Question 14.
If the radius of the octahedral void is r and radius of the atoms in close packing is R, derive relation between r and R.
Answer:

A sphere fitting into the octahedral void is shown by shaded circle. A sphere above and a sphere below this small (shaded) sphere have not been shown in the figure. Obviously, ABC is a right angled triangle. Applying Pythagoras theorem,
BC² = AB² + BC²

Question 15.
Copper crystallises into a fee lattice with edge length 3.61 x 10^-8 cm. Show that the calculated density is in agreement with its measured value of 8.92 g cm^-3.
Answer:

which is in close agreement with the measured value.

Question 16.
Analysis shows that nickel oxide has the formula Ni₀₉₈ O_1.00 What fractions of nickel exist as Ni²⁺ and Ni³⁺ ions?
Answer:
The formula Ni₀₉₈ O_1.00 shows that Ni : O= 0.98 : 1.00 = 98 : 100.
Thus, if there are 100 O-atoms, then Ni atoms = 98
charge on 100 O2′ ions = 100 x (-2) = – 200
suppose Ni atoms as Ni²⁺ = x
Then Ni atoms as Ni³⁺ = 98 – x
Total charge on Ni2+ and Ni³⁺
= (+2) x + (+3) (98 – x) = 294 – x
As metal oxide is neutral, total charge on
cations = total charge on anions
294 – x = 200
x = 94
∴ % of Ni as Ni²⁺ =
gxlOO
98
= 96% % of Ni as Ni³⁺ = 1000 – 96 = 4.

Question 17.
What is a semiconductor? Describe the two main types of semiconductors and contrast their conduction mechanism.
Answer:
Substances whose conductance lies in between that of metals (conductors) and insulators are called semiconductors. Two main types of semiconductors are n-type and p-type. n-type semiconductor: When a silicon crystal is doped with atoms of group-15 elements, such as P, As, Sb or Bi, then only four of the five valance electrons of each impurity atom participate in forming covalent bonds and fifth electron is almost free to conduct electricity. Silicon that has been doped with group-15 element is called n-type semiconductor.

(ii) p-type semiconductor: when a silicon crystal is doped with atoms of group-13 elements, such as B, Al, Ga or In, each impurity atom forms only three covalent bonds with the host atoms. The place where the fourth electron is missing t is called a hole which moves through the crystal like a positive charge and hence increases its conductivity. Silicon that has been doped with group-13 element is p-type semiconductor.

Question 18.
Non-stoichiometric cuprous oxide, Cu₂O can be prepared in laboratory. In this oxide, copper to oxygen ratio is slightly less than 2:1. Can you account for the fact that this substance is a p-type semiconductor?
Answer:
The ratio less than 2:1 in Cu₂O shows that some cuprous (Cu⁺) ions have been replaced by cupric (Cu²⁺) ions. To maintain electrical neutrality, every two Cu⁺ ions will be replaced by one Cu²⁺ ion thereby creating a hole. As conduction will be due to presence of these positive holes, hence it is a p-type semiconductor.

Question 19.
Ferric oxide crystallises in a hexagonal close-packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions. Derive the formula of the ferric oxide.
Answer:
O^2- ions form cubic close packed lattice
Number of O^2- = 8 x \(\frac { 1 }{ 8 }\) = 1
Number of Fe³⁺ = \(\frac { 2 }{ 3 }\) octahedral voids = \(\frac { 2 }{ 3 }\)
Formula of the compound =

Question 20.
Classify each of the following as being either a p-type or a n-type semiconductor: (i) Ge. doped with In (ii) B doped with Si.
Answer:
(i) Ge is group 14 element and In is group 13 element. Hence, an electron deficit hole is created and therefore, it is p-type.
(ii) B is group 13 element and Si is group 14 element, there will be free electron. Hence it is p-type.

Question 21.
Gold (atomic radius = 0.144 nm) crystallises in a face-centred unit cell. What is the length of a side of the cell?
Answer:
For fee lattice, edge length,
a = 2 √2 x 0.144 nm = 0.407 nm

Question 22.
In terms of band theory, what is the difference
(i) between a conductor and an insulator
(ii) between a conductor and a semi¬conductor?
Answer:
(i) The energy gap between the valence band and conduction band in an insulator is very large whereas in a conductor, the energy gap is very small or there is overlapping between valance band and conduction band.
(ii) In a conductor, there is very small energy gap or there is overlapping between valence band and conduction band but in a semiconductor there is always a small energy gap between them. ’

Question 23.
Explain the following terms with suitable examples:
(i) Schottky defect
(ii) Frenkel defect
(iii) Interstitials and
(iv) F-centres.
Answer:
(i) Schottky defect:
It arises due to the missing of equal number of cations and anions from their normal positions leaving behind a pair of holes. This defect is usually observed in ionic compounds having high coordination number and ions of almost similar size. The density of the crystal is lowered due to the presence of vacancies in the crystal lattice. Schottky defect increases slightly the electrical conductance of the crystal. E.g. NaCl, KC1, CsCl.

(ii) Frenkel defect:
It arises w’hen an ion, usually cation, leaves its normal site and occupies an interstitial site. This defect is usually observed in ionic ‘ compounds having low co-ordination number , and crystals with anions are much larger in size than the cations. Since no ions are missing from the crystals as whole, it doesn’t affect the density of the crystal.
E.g.: AgCl, ZnS, AgBr

(iii) Interstitials:
Atoms or ions which occupy the normally vacant interstitial positions in a crystal are called interstitials.

(iv) F-centres:
The electron occupying holes, created by missing of anions from the lattice sites are called F-centres. These F-centres are responsible for colour of compound.

Question 24.
Aluminium crystallises in a cubic close- packed structure. Its metallic radius is 125 pm.
(i) What is the length of the side of the unit cell?
(ii) How many unit cells are there in 1.00 cm3 of aluminium?
Answer:
(i) For CCP structure, edge length, a = 2√2r = 2√2 x 125pm = 354pm
(ii) volume of unit cell = a³ = (354 × 10^-8m)³
∴ Number of unit cells in 1.00 cm³

Question 25.
If NaCl is doped with 10^-3 mol% of SrCl₂, what is the concentration of cation vacancies?
Answer:
Introduction of one Sr²⁺ introduces one cation vacancy because Sr²⁺ replaces two Na⁺ ions. Introduction of 10^-3 moles of SrCl₂ per 100 moles of NaCl would introduce 10’3 mole cation vacancies
∴ Number of vacancies per mole of NaCl
= \(\frac{10^{-3}}{100}\) = 10^-5 mole = 10^-5 x 6.02 x 10²³ 1 (F = 6.02 x 10¹⁸ vacancies.

Question 26.
Explain the following with suitable examples:
(i) Ferromagnetism
(ii) Paramagnetism
(iii) Ferrimagnetism
(iv) Antiferromagnetism
(v) 12 – 16 and 13 – 15 group compounds.
Answer:
(i) Ferromagnetism is considered as an extreme case of paramagnetism. Those substances with a strong attraction to an external magnetic field and those retain a permanent magnetism even when the field is removed are known as ferromagnetic substances. Ferrimagnetism is caused by spontaneous alignment of magnetic moment of atoms or ions in the same direction. Examples of ferromagnetic substances are Fe, CO, Ni, CrO₂ alloys of Fe, Co and Ni etc. Once such a material is, magnetised, it remains permanently magnetised Ferromagnetic

(ii) Paramagnetic substances are those which are attracted by magnetic field and have impaired electrons. They lose magnetism in the absence of magnetic field. The greater the number of unpaired electrons, the greater the magnetic moment of the substance and hence greater the paramagnetism. E.g.: O₂, CuO, Fe³⁺, Al, Mn, Cu²⁺ etc.

(iii) These are substances which are expected to have large magnetism on the basis of unpaired electrons but actually have small net magnetic moment. This is due to the alignment of magnetic moment in opposite direction in unequal numbers resulting in a net magnetic moment.

(iv) These are substances which are expected to possess paramagnetism or ferromagnetism on the basis of unpaired electrons but they actually possess zero net magnetic moment. It arises due to the alignment of magnetic moments in opposite direction in a compensatory manner resulting a zero magnetic moment.

(v) Germanium and silicon are group 14 elements and have a characteristic valance of four and form four bonds as in diamond. A large variety of solid state materials have been prepared by combination of groups 13 and 15 or 12 and 16 to stimulate average valence of four as Ge or Si. Typical compounds of group 13 -15 are InSb, A1P and GaAs. Gallium arsenide (GaAs) semiconductors have very fast response and have revelutionised semi – conductor devices. ZnS, CdS, CdSe and HgTe are examples of group 12 – 16 compounds. In. these compounds, the bonds are not perfectly covalent and the ionic character depends on the electronegativities of the two elements.

2nd PUC Chemistry The Solid State One Mark Questions and Answers

Question 1.
How many atoms are present in the unit
cell of
(a) simple cubic lattice
(b) BCC
(c) FCC
Answer:
(a) 1
(b) 2 and
(c) 4.

Question 2.
How can you convert NaCl structure into CsCl structure and vice versa.
Answer:
NaCl structure can be converted into CsCl. structure by application of pressure while the reverse can be done by heating to 760 k.

Question 3.
How does amorphous silica differ from quartz?
Answer:
In amorphous Silica, SiO₄ tetrahedra are randomly joined to each other whereas in quartz, they are linked in a regular manner.

Question 4.
What are stoichiometric defects or intrisnic defects in ionic crystals?
Answer:
Stoichiometric defects are those defects in which the ratio of cations to anions remains the same are represented by the molecular formula

Question 5.
What is the effect of the presence of Schottky defects on the density of a crystal?
Answer:
The density of the crystal decreases.

Question 6.
Name one Solid which has both Schottky and Frenkel defects. !
Answer:
Silver bromide (AgBr).

Question 7.
What is Frenkel, defect?
Answer:
When some ions (usually cations) are missing from the lattice sites and they occupy the interstitial sites so that electrical neutrality as well as stoichiometric ratio is maintained, it is called Frenkel defect.

Question 8.
Why does the conductivity of metals decrease with rise of temperature?
Answer:
The kernals begin to vibrate in the path of the flow of electrons creating hinderance in the flow.

Question 9.
What are the type of lattice imperfections found in crystals?
Answer:
(a) Stiochiometric defects, viz, Schottky defect and Frenkel defect and
(b) Non- Stiochiometric defects, viz., metal excess, metal deficiency and
(c) Impurity defects.

Question 10.
Define coordination number of a metal ion in an ionic crystal.
Answer:
In an ionic crystal, coordination number of a metal ion (+ve ion) is the number of negative ions surrounding the metal ion, i.e. which are present as its nearest neighbours.

Question 11.
Why are amorphous solids to be considered as supercooled liquids?
Answer:

• Like liquids, amorphous solids are isotropic
• Like liquids, they posses fluidity.

2nd PUC Chemistry The Solid State Two Marks Questions and Answers

Question 1.
Refractive index of a solid is observed to have the same value along all directions. Comment on the nature of solid. Would it show cleavage property?
Answer:
As the solid has same value of refractive index along all directions. This means that it is isotropic and hence amorphous. Being an amorphous solid, it would not show a clean cleavage when cut with a knife. Instead it would
break into pieces with irregular surfaces.

Question 2.
Ionic solids conduct electricity in molten state but not in solid state. Explain.
Answer:
In molten state, ionic solids dissociate to give free ions and hence can conduct electricity.

However, in the solid state, as the ions are not free but remain held together by strong electrostatic forces of attraction, they cannot
conduct electricity in the solid state.

Question 3.
Name the parameters that characterize a unit cell.
Answer:
A unit cell is characterized by following six parameters: (i) The dimensions of the unit cell along the three edges. These are represented by a, b and c. The edges may or may not be mutually perpendicular. (ii) The angles between the edges. These are represented by α, β  and γ. The angle α  is. between b and c, β is between a and c and γ
is between a and b.

Question 4.
What type of defect can arise when a solid is heated? Which physical property is effected and in what way?
Answer:
When a solid is heated, vacancy defect is produced in the crystal. This is because of heating, some atoms or ions leave the lattice site completely, i.e. some lattice site becomes vacant. As a result of this defect, the density of the substance decreases because some atoms/ions leave the crystal completely.

Question 5.
A group 14 element is to be converted into n-type semiconductor by doping it with a suitable impurity. To which group should the impurity belong?
Answer:
n-type semiconductor means conduction due to presence of excess of negatively charged electrons. Hence to convert group 14 element into n-type semiconductor, it should be doped with group 15 element.

Tili Kannada Text Book Class 9 Rachana Bhaga Gadegala Vistarane