NCERT Solutions Class 11 Physics Chapter 12 Thermodynamics

Question :1.
A geyser heats water flowing at the rate of 3.0 litres per minute from
(27 °C to 77 °C) If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is 4.0 × 10⁴ J/g?

Answer :
Given:
Water is flowing at a rate of = 3.0 litre/min
The geyser heats the water, raising the temperature from 27°C to 77°C.
Initial temperature, T₁ = 27°C
Final temperature, T₂ = 77°C
∴Rise in temperature, ΔT = (T₂-T₁)
            = (77 – 27) = 50°C
Heat of combustion = (4 × 10⁴) J/g°C
Specific heat of water, c = 4.2 J^-1 g^-1°C^-1
Mass of flowing water, m = 3.0 litre/min = 3000 g/min
Total heat used, ΔQ = (mc ΔT)
= 3000 × 4.2 × 50
= (6.3 × 10⁵) J/min
∴Rate of consumption = (6.3×10⁵)/ (4×10⁴) = 15.75g/min

Question : 2.
What amount of heat must be supplied to 2.0 × 10^–2 kg of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure?
(Molecular mass of N₂ = 28; R = 8.3 J mol^–1 K^–1.)

Answer :
Given:
Mass of nitrogen, m = (2.0 × 10^–2) kg = 20 g
Rise in temperature, ΔT = 45°C
Molecular mass of N₂, M = 28
Universal gas constant, R = 8.3 J mol^–1 K^–1
Number of moles, n = (m/M)
= (2.0×10^-2x10³)/ (28) = 0.714
Molar specific heat at constant pressure for nitrogen = C_p = (7/2) R
= (7)/ (2×8.3) = 29.05 J mol^–1 K^–1
The total amount of heat to be supplied is given by the relation:
ΔQ = (nC_P ΔT)
= 0.714 × 29.05 × 45
= 933.38 J
Therefore, the amount of heat to be supplied is 933.38 J.

Question : 3.
Explain why
(a) Two bodies at different temperatures T₁ and T₂ if brought in thermal contact do not necessarily settle to the mean temperature (T₁ + T₂)/2.
(b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat.
(c) Air pressure in a car tyre increases during driving.
(d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude.

Answer :

• When two bodies at different temperatures T₁ and T₂ are brought in thermal contact, heat flows from the body at the higher temperature to the body at the lower temperature till equilibrium is achieved, i.e., the temperatures of both the bodies become equal. The equilibrium temperature is equal to the mean temperature ((T₁ + T₂)/2) only when the thermal capacities of both the bodies are equal.
• The coolant in a chemical or nuclear plant should have a high specific heat. This is because higher the specific heat of the coolant, higher is its heat-absorbing capacity and vice versa. Hence, a liquid having a high specific heat is the best coolant to be used in a nuclear or chemical plant. This would prevent different parts of the plant from getting too hot.
• When a car is in motion, the air temperature inside the car increases because of the motion of the air molecules. According to Charles’ law, temperature is directly proportional to pressure. Hence, if the temperature inside a tyre increases, then the air pressure in it will also increase.
• A harbour town has a more temperate climate (i.e., without the extremes of heat or cold) than a town located in a desert at the same latitude. This is because the relative humidity in a harbour town is more than it is in a desert town.

Question : 4.
A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure.
The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it.
By what factor does the pressure of the gas increase if the gas is compressed to half its original volume?

Answer :
The cylinder is completely insulated from its surroundings. As a result, no heat is exchanged between the system (cylinder) and its surroundings. Thus, the process is adiabatic.
Initial pressure inside the cylinder = P₁
Final pressure inside the cylinder = P₂
Initial volume inside the cylinder = V₁
Final volume inside the cylinder = V₂
Ratio of specific heats, γ = 1.4
For an adiabatic process, we have: (P₁V₁^γ) = (P₂V₂^γ)
The final volume is compressed to half of its initial volume.
V₂= (V₁/2)
(P₁V₁^γ) = (P₂ (V₁/2) ^γ)
 = (P₂/P₁) = (V₁^γ)/ ((V₁/2) ^γ) = (2) ^γ = (2)^1.4 =2.639
Hence, the pressure increases by a factor of 2.639.

Question : 5.
In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system.
If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal,
how much is the net work done by the system in the latter case? (Take 1 cal = 4.19 J)

Answer :
The work done (W) on the system while the gas changes from state A to state B is 22.3 J.
This is an adiabatic process. Hence, change in heat is zero.
ΔQ = 0
ΔW = –22.3 J (Since the work is done on the system)
From the first law of thermodynamics, we have:
ΔQ = (ΔU + ΔW)
Where,
ΔU = Change in the internal energy of the gas
ΔU = (ΔQ – ΔW) = – (– 22.3 J)
ΔU = + 22.3 J
When the gas goes from state A to state B via a process, the net heat absorbed by the system is:
ΔQ = 9.35 cal
= (9.35 × 4.19)
 = 39.1765 J
 Heat absorbed, ΔQ = (ΔU + ΔQ)
ΔW = (ΔQ – ΔU)
= 39.1765 – 22.3
= 16.8765 J
Therefore, 16.88 J of work is done by the system.

Question : 6.
Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated.
The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following:
(a) What is the final pressure of the gas in A and B?
(b) What is the change in internal energy of the gas?
(c) What is the change in the temperature of the gas?
(d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface?

Answer :
(a) When the stopcock is suddenly opened, the volume available to the gas at 1 atmospheric pressure will become two times.
Therefore, pressure will decrease to one-half, i.e., 0.5 atmospheres.
(b) There will be no change in the internal energy of the gas as no work is done on/by the gas.
(c) Since no work is being done by the gas during the expansion of the gas, the temperature of the gas will not change at all.
(d) No, because the process called free expansion is rapid and cannot be controlled.
The intermediate states are non-equilibrium states and do not satisfy the gas equation. In due course, the gas does return to an equilibrium state.

Question : 7.
A steam engine delivers 5.4×10⁸ J of work per minute and services 3.6 × 10⁹J of heat per minute from its boiler.
What is the efficiency of the engine? How much heat is wasted per minute?

Answer :
Work done by the steam engine per minute, W = (5.4 × 10⁸) J
Heat supplied from the boiler, H = (3.6 × 10⁹) J
Efficiency of the engine = (Output energy)/ (Input energy)
η = (W/H)
= (5.4 x 10⁸) / (3.6 × 10⁹) = 0.15
Hence, the percentage efficiency of the engine is 15 %.
Amount of heat wasted = (3.6 × 10⁹) – (5.4 × 10⁸)
= (30.6 × 10⁸)
 = 3.06 × 10⁹ J
Therefore, the amount of heat wasted per minute is 3.06 × 10⁹ J.