**NCERT Solutions for Class 11 Physics Physics Chapter 12 Thermodynamics** includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Physics exams must go through NCERT Solutions for **Class 11 Physics Physics Chapter 12 Thermodynamics**. Going through the solutions provided on this page will help you to know how to approach and solve the problems.

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## NCERT Solutions for Class 11 Physics Physics Chapter 12 Thermodynamics

NCERT Solutions for Class 11 Physics Physics Chapter 12 Thermodynamics are been solved by expert teachers of CBSETuts.com. All the solutions given in this page are solved based on CBSE Syllabus and NCERT guidelines.

**NCERT Exercises
**

**Question 1.**

A geyser heats water flowing at the rate of 3.0 liters 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 x 10^{4} J g^{_1} ?

**Answer:**

Here, mass of water heated,

m = 3000 g min^{-1
}Volume of water heated = 3.0 lit min^{1
}Rise in temperature ΔT = (77 – 27)°C = 50°C

Specific heat of water c = 4.2 J s^{-1} °C^{-1}

Amount of heat used ΔQ = mc ΔT

**Question 2.**

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

**Answer:**

Mass of nitrogen gas,

m = 2 x 10^{2} kg = 20 g

Molecular mass of nitrogen gas = 28 g

Rise in temperature, ΔT = 45°C

Heat supplied = ?

\(n=\cfrac { m }{ 2 } =\cfrac { 20 }{ 28 } =\cfrac { 5 }{ 7 }\)

For nitrogen (N_{2}) which is a diatomic gas, molar specific heat at constant pressure

C_{p} = \(\cfrac { 7 }{ 2 }\) R where R = 8.3 J mol^{-1} K^{_1}^{
}Therefore, heat supplied to the gas

ΔQ = nC_{p}ΔT =\(\cfrac { 5 }{ 7 }\) x \(\cfrac { 7 }{ 2 }\) x R x 45 = \(\cfrac { 5 }{ 7 }\) x \(\cfrac { 7 }{ 2 }\)x8.3 x 45

ΔQ = 933.8 J = 934 J

**Question 3.**

Explain why

**(a)** Two bodies at different temperatures T_{1}) and T_{2} if brought in thermal contact do not necessarily settle to the mean temperature (T_{1} + T_{2})/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:**

**(a)** In thermal contact, heat flows from the body at higher temperature to the body at lower temperature till temperatures become equal. The final temperature can be the mean temperature (T_{1}+ T_{2})/2 only when thermal capacities of the two bodies are equal.

**(b)** This is because heat absorbed by a substance is directly proportional to the specific heat of the substance.

**(c)** During driving, the temperature of air inside the tyre increases due to motion. According to Charle’s law, P T. Therefore, air pressure inside the tyre increases.

**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:**

As the gas is completely insulated, the process is adiabatic.

**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:**

ln first case, the state of the gas changes adiabatically from A to

Therefore, ΔQ = 0, ΔW= – 22.3 J

If ΔH is change in internal energy of the system then,

ΔQ = ΔH + ΔW

0 = ΔH + ΔW

ΔH = – ΔW = – (- 22.3 J) = 22.3 J

In the second case, when the state A is taken to state B, the heat absorbed by the system

ΔQ = 9.35 cal = 9.35 x 4.2 J = 39.27 J = 39.3 J

ΔW = ?

Applying first law of thermodynamics ΔQ = ΔH + ΔW

ΔW =ΔH – ΔQ = 39.3 – 22.3 = 17.0 J

**Question 6.**

Two cylinders A and B of equal capacity are connected to each other via a stopcock. The cylinder A contains a gas at standard temperature and pressure, while the cylinder 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 setting 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 atmosphere pressure will become two times. Therefore, pressure will decrease to one-half,i.e.. 0.5 atmosphere.

**(b)** There will be no change in the internal energy of the gas as no work is done on/ by the gas.

**(c)** Also, there will be no change in temperature of the gas as gas does no work in expansion.

**(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 x 10^{8} J of work per minute and services 3.6 x 10^{9} J of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?

**Answer:**

Useful work done per min (output)

= 5.4 x 10^{8} J

Heat absorbed per min (input) = 3.6 x 10^{9} J

= 0.15 = 0.15 x 100% = 15%

Heat energy wasted per minute = Heat energy

wasted per useful work done per minute

= 3.6 x 10^{9} – 5.4 x 10^{8} = 3.06 x 10^{9} J min^{-1}

**Question 8.**

An electric heater supplies heat to a system at a rate of 100 W. If system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing?

**Answer:**

Given that

Heat supplied ΔQ = 100 W = 100 J s

Useful work done ΔW = 75 J s^{-1
}Change internal energy ΔU = ?

According to first law of thermodynamics,

change in internal energy is given by

ΔU = ΔQ – ΔW

ΔU = 100 – 75 = 25 J s-^{1} = 25 W

**Question 9.**

A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in figure.

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.

**Answer:**

Total work done by gas equals to area enclosed.

W = Area DEED

= \(\cfrac { 1 }{ 2 }\) x EF x DF

= \(\cfrac { 1 }{ 2 }\) x (5 – 2) x (600 – 300) =450J

**Question 10.**

A refrigerator is to maintain eatables kept inside at 9°C. If room temperature is 36°C, calculate the coefficient of performance.

**Answer:**

Here, T_{1} = 36°C = 36 + 273 = 309 K

T_{2} = 9°C = 9 + 273 = 283 K

Coefficient of performance of the refrigerator

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