Page No 296: - Chapter 11 Thermal Properties Of Fluids class 11 ncert solutions Physics - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question 11.15:

Given below are observations on molar specific heats at room temperature of some common gases.

The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic 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:

The gases listed in the given table are diatomic. Besides the translational degree of freedom, they have other degrees of freedom (modes of motion).

Heat must be supplied to increase the temperature of these gases. This increases the average energy of all the modes of motion. Hence, the molar specific heat of diatomic gases is more than that of monatomic gases.

If only rotational mode of motion is considered, then the molar specific heat of a diatomic 

With the exception of chlorine, all the observations in the given table agree with  . This is because at room temperature, chlorine also has vibrational modes of motion besides rotational and translational modes of motion.

Question 11.16:

Answer the following Questions based on the P–T phase diagram of carbon dioxide:

(a) At what temperature and pressure can the solid, liquid and vapour phases of CO₂ co-exist in equilibrium?

(b) What is the effect of decrease of pressure on the fusion and boiling point of CO₂?

(c) What are the critical temperature and pressure for CO₂? What is their significance?

(d) Is CO₂ solid, liquid or gas at (a) –70 °C under 1 atm, (b) –60 °C under 10 atm, (c) 15 °C under 56 atm?

Answer:

(a) The P–T phase diagram for CO₂ is shown in the following figure.

C is the triple point of the CO₂ phase diagram. This means that at the temperature and pressure corresponding to this point (i.e., at –56.6°C and 5.11 atm), the solid, liquid, and vaporous phases of CO₂ co-exist in equilibrium.

(b) The fusion and boiling points of CO₂ decrease with a decrease in pressure.

(c) The critical temperature and critical pressure of CO₂ are 31.1°C and 73 atm respectively. Even if it is compressed to a pressure greater than 73 atm, CO₂ will not liquefy above the critical temperature.

(d) It can be concluded from the P–T phase diagram of CO₂ that:

(a) CO₂ is gaseous at –70°C, under 1 atm pressure

(b) CO₂ is solid at –60°C, under 10 atm pressure

(c) CO₂ is liquid at 15°C, under 56 atm pressure

Question 11.17:

Answer the following Questions based on the P–T phase diagram of CO₂:

(a) CO₂ at 1 atm pressure and temperature – 60 °C is compressed isothermally. Does it go through a liquid phase?

(b) What happens when CO₂ at 4 atm pressure is cooled from room temperature at constant pressure?

(c) 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.

(d) CO₂ is heated to a temperature 70 °C and compressed isothermally. What changes in its properties do you expect to observe?

Answer:

Answer:

(a) No

(b) It condenses to solid directly.

(c) The fusion and boiling points are given by the intersection point where this parallel line cuts the fusion and vaporisation curves.

(d) It departs from ideal gas behaviour as pressure increases.

Explanation:

(a) The P–T phase diagram for CO₂ is shown in the following figure.

At 1 atm pressure and at –60°C, CO₂ lies to the left of –56.6°C (triple point C). Hence, it lies in the region of vaporous and solid phases.

Thus, CO₂ condenses into the solid state directly, without going through the liquid state.

(b) At 4 atm pressure, CO₂ lies below 5.11 atm (triple point C). Hence, it lies in the region of vaporous and solid phases. Thus, it condenses into the solid state directly, without passing through the liquid state.

(c) When the temperature of a mass of solid CO₂ (at 10 atm pressure and at –65°C) is increased, it changes to the liquid phase and then to the vaporous phase. It forms a line parallel to the temperature axis at 10 atm. The fusion and boiling points are given by the intersection point where this parallel line cuts the fusion and vaporisation curves.

(d) If CO₂ is heated to 70°C and compressed isothermally, then it will not exhibit any transition to the liquid state. This is because 70°C is higher than the critical temperature of CO₂. It will remain in the vapour state, but will depart from its ideal behaviour as pressure increases.

Question 11.18:

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 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:

Initial temperature of the body of the child, T₁ = 101°F

Final temperature of the body of the child, T₂ = 98°F

Change in temperature, ΔT =   °C

Time taken to reduce the temperature, t = 20 min

Mass of the child, m = 30 kg = 30 × 10³ g

Specific heat of the human body = Specific heat of water = c

= 1000 cal/kg/ °C

Latent heat of evaporation of water, L = 580 cal g^–1

The heat lost by the child is given as:

Let m₁ be the mass of the water evaporated from the child’s body in 20 min.

Loss of heat through water is given by:

∴Average rate of extra evaporation caused by the drug 

Question 11.19:

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 of ice is put in the box, estimate the amount of ice remaining after 6 h. The outside temperature is 45 °C, and co-efficient of thermal conductivity of thermacole is 0.01 J s^–1 m^–1 K^–1. [Heat of fusion of water = 335 × 10³ J kg^–1]

Answer:

Side of the given cubical ice box, s = 30 cm = 0.3 m

Thickness of the ice box, l = 5.0 cm = 0.05 m

Mass of ice kept in the ice box, m = 4 kg

Time gap, t = 6 h = 6 × 60 × 60 s

Outside temperature, T = 45°C

Coefficient of thermal conductivity of thermacole, K = 0.01 J s^–1 m^–1 K^–1

Heat of fusion of water, L = 335 × 10³ J kg^–1

Let m^’ be the total amount of ice that melts in 6 h.

The amount of heat lost by the food:

Where,

A = Surface area of the box = 6s² = 6 × (0.3)² = 0.54 m³

Mass of ice left = 4 – 0.313 = 3.687 kg

Hence, the amount of ice remaining after 6 h is 3.687 kg.

Question 11.20:

A brass boiler has a base area of 0.15 m² and thickness 1.0 cm. It boils 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 ^–1 m^–1 K^–1; Heat of vaporisation of water = 2256 × 10³ J kg^–1.

Answer:

Base area of the boiler, A = 0.15 m²

Thickness of the boiler, l = 1.0 cm = 0.01 m

Boiling rate of water, R = 6.0 kg/min

Mass, m = 6 kg

Time, t = 1 min = 60 s

Thermal conductivity of brass, K = 109 J s ^–1 m^–1 K^–1

Heat of vaporisation, L = 2256 × 10³ J kg^–1

The amount of heat flowing into water through the brass base of the boiler is given by:

Where,

T₁ = Temperature of the flame in contact with the boiler

T₂ = Boiling point of water = 100°C

Heat required for boiling the water:

θ = mL … (ii)

Equating equations (i) and (ii), we get:

Therefore, the temperature of the part of the flame in contact with the boiler is 237.98°C.