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NCERT Class 11 Physics Chapter 11 Thermodynamics
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Thermodynamics
Chapter: 11
| Part – II |
EXERCISE
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?
Ans: Energy required to heat the water:
Mass of water
= 3.0 litres/minute × 1000 g/litre
= 3000 g/minute
Heat capacity of water = 4.184 J/g°C
Temperature change = 77°C – 27°C = 50°C
Energy = 3000 g/minute × 4.184 J/g°C × 50°C
= 627600 J/minute
Energy into joules per second (J/s):
627600 J/minute ÷ 60 s/minute = 10460 J/s
Heat of combustion = 4.0 × 104 J/g
Rate of fuel consumption:
= 10460 J/s ÷ 4.0 × 104 J/g
= 0.2615 g/s or 15.69 g/minute.
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.)
Ans: Here:
Q = n × Cp × ΔT
n = mass / molecular mass
= 2.0 × 10–2 kg / 28 kg/kmol
= 7.14 × 10–4 kmol
Cp = (7/2) × R
= (7/2) × 8.3 J/mol·K
= 29.05 J/mol·K
Q = 7.14 × 10–4 kmol × 29.05 J/mol·K × 45 K
= 93.5 J.
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.
Ans: Law of conservation of Energy says that the energy is transferred from one body to another and from one format to another. The final temperature can be the mean temperature only when the thermal capacities of both the bodies are equal.
(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.
Ans: If the specific heat of the coolant is high, it can absorb large amounts of heat without heating itself much. Hence, a coolant should have high specific heat.
(c) Air pressure in a car tyre increases during driving.
Ans: Temperature of air in the tyre increases due to friction of tyre with road. Therefore, air pressure inside the tyre increases according to Charle’s law.
(d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude.
Ans: This is because in a harbour town, the relative humidity is more than in a desert town. Hence, the climate of harbour town is without extremes of hot and cold.
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 ab pressure of the gas increase if the gas is compressed to half its original volume?
Ans: Here:
P1V1γ = P2V2γ
where γ is the adiabatic index (approximately 1.4 for hydrogen).
Given:
V2 = V1/2
Rearrange the equation to solve for P2/P1:
P2/P1 = (V1/V2)γ
= (2)1.4
= 2.639
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)
Ans: Here:
9.35 cal × 4.19 J/cal = 39.15 J
The first law of thermodynamics:
ΔU = Q – W
For the adiabatic process:
ΔU = 0 – (-22.3 J)
ΔU = 22.3 J
For the non-adiabatic process:
ΔU = 39.15 J – W
Since ΔU is the same for both processes:
22.3 J = 39.15 J – W
W = 16.85 J.
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 suddenly opened. Answer the following:
(a) What is the final pressure of the gas in A and B?
Ans: Here:
P2V2 = P1V1
P1 = 1 atm,
V1 = V
V2 = 2V and P2 = ?
P2 = P1V1/V2
= 1 × V/ 2V
= 0.5 atm.
(b) What is the change in internal energy of the gas?
Ans: Since the temperature of the system remains unchanged, and the process is adiabatic, the change in internal energy of the gas is zero.
(c) What is the change in the temperature of the gas?
Ans: The system being internally insulated, there is no change in temperature.
(d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface?
Ans: No, because the process called free expansion is rapid and cannot be controlled. Therefore, the intermediate states are non – equilibrium states and the gas equation is not satisfied in these states. The gas cannot return to an equilibrium state which lie on the P – T- V surface.
7. An electric heater supplies heat to a system at a rate of 100W. If system performs work at a rate of 75 joules per second. At what rate is the internal
energy increasing?
Ans: The first law of thermodynamics states:
dU/dt = dQ/dt – dW/dt
Given:
dQ/dt = 100 W
dW/dt = 75 J/s
dU/dt = 100 W – 75 J/s
= 25 J/s.
8. A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig. (11.13).
Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the total work done by the gas from D to E to F.
Ans: As it is clear from the above figure, Change in pressure, dp = EF.
Area of triangle = (1/2) × base x height
Base = 5.0 m³ – 2.0 m³ = 3.0 m³
Height = 600 N/m² – 300 N/m² = 300 N/m²
Work done from D to E = (1/2) × 3.0 m³ × 300 N/m² = 450 J.
Work done from E to F:
Length = 5.0 m3 – 2.0 m3 = 3.0 m3
Width = 300 N/m2
Work done from E to F = 3.0 m3 × 300 N.m2 = 900 J
Total work done
= 450 j + 900 j
= 1350 J.
