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NCERT Exemplar Class 11 Physics Solutions Chapter 12 Thermodynamics

NCERT Exemplar Class 11 Physics Solutions Chapter 12 Thermodynamics

Edited By Vishal kumar | Updated on Apr 09, 2025 11:20 AM IST

Ever curious about how heat moves, why engines function, or why a hot cup of tea gets cold after some time? Thermodynamics is the answer to all these interesting phenomena!

In NCERT Exemplar Class 11 Physics Solutions Chapter 12, students will learn about concepts such as heat, work, internal energy, and thermodynamic laws. This chapter is the gateway to understanding practical applications, ranging from refrigerators to automobile engines. Get started to learn these crucial concepts with step-by-step solutions!

This Story also Contains
  1. NCERT Exemplar Class 11 Physics Solutions Chapter 12: MCQI
  2. NCERT Exemplar Class 11 Physics Solutions Chapter 12: MCQII
  3. NCERT Exemplar Class 11 Physics Solutions Chapter 12: Very Short Answer
  4. NCERT Exemplar Class 11 Physics Solutions Chapter 12: Short Answer
  5. NCERT Exemplar Class 11 Physics Solutions Chapter 12: Long Answer
  6. NCERT Exemplar Class 11 Physics Solutions Chapter 12 Thermodynamics: Topics
  7. Important Concepts & Formulas - NCERT Exemplar Class 11 Physics Chapter 12 (Thermodynamics)
  8. NCERT Exemplar Class 11 Physics Solutions Chapter-Wise
  9. NCERT Exemplar Class 11th Solutions
NCERT Exemplar Class 11 Physics Solutions Chapter 12 Thermodynamics
NCERT Exemplar Class 11 Physics Solutions Chapter 12 Thermodynamics

NCERT Exemplar Class 11 Physics Solutions Chapter 12 deals with significant topics of heat, energy, and work. This chapter discusses how heat is transferred to energy and how it influences matter. Students will study the laws of thermodynamics and their uses in real life. These solutions make it easy to solve exam questions and understand significant formulas for better preparation.

Background wave

NCERT Exemplar Class 11 Physics Solutions Chapter 12: MCQI

Question:1

An ideal gas undergoes four different processes from the same initial state (Figure). Four processes are adiabatic, isothermal, isobaric and isochoric. Out of 1, 2, 3 and 4 which one is adiabatic.

P - V graph
(a) 4
(b) 3
(c) 2
(d) 1

Answer:

The answer is the option (c) 4 is the isobaric process, 1 is isochoric. Out of 3 and 2, 3 has a smaller slope hence is isothermal. Remaining process 2 is adiabatic.

Question:2

If an average person jogs, he produces 14.5×103cal/min. This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming 1 kg requires 580×103call for evaporation) is
(a) 0.025 kg
(b) 2.25 kg
(c) 0.05 kg
(d) 0.20 kg

Answer:

The answer is the option (a) 0.025 kg

Amount of sweat produced=14.5×103580×103=0.025Kg

Question:3

Consider P-V diagram for an ideal gas shown in Figure.
P - V graph for ideal gas
Out of the following diagrams (Figure), which represents the T-P diagram?

Time - Pressure Graph
(a) (iv)
(b) (ii)
(c) (iii)
(d) (i)

Answer:

The answer is the option (c)

Question:4

An ideal gas undergoes cyclic process ABCDA as shown in given P-V diagram (Figure). The amount of work done by the gas is
P - V graph for cyclic process
a) 6P0V0
b) 2P0V0
c) +2P0V0
d) +4P0V0

Answer:

The answer is the option (b). Since the direction of arrows is anticlockwise, so work done is negative equal to the area of the loop
=2P0V0

Question:5

Consider two containers A and B containing identical gases at the same pressure, volume and temperature. The gas in container A is compressed to half of its original volume isothermally while the gas in container B is compressed to half of its original value adiabatically. The ratio of final pressure of gas in B to that of gas in A is
a) 2γ1
b) (12)γ1
c) (11γ)2
d) (1γ1)2

Answer:

The answer is the option (a). Let us consider a P-V diagram for container A and B. Compression of gas is involved in both the cases. For the isothermal process (gas A) during 12

P1V1=P2V2

P0(2V0)=P2V0

P2=2P0
For adiabatic process 12
P1V1γ=P2V2γ

P0(2V0)γ=P2(V0)γ

P2=(2V0V0)γP0=2γP0

(P2)B(P1)A=2γP02P0=2γ1

Question:6

Three copper blocks of massesM1,M2andM3 kg respectively are brought into thermal contact till they reach equilibrium. Before contact, they were at T1,T2,T3(T1>T2>T3). Assuming there is no heat loss to the surroundings, the equilibrium temperature T is (s is specific heat of copper)

a) T=T1+T2+T33
b) T=M1T1+M2T2+M3T3M1+M2+M3
c) T=M1T1+M2T2+M3T33(M1+M2+M3)
d) T=M1T1s+M2T2s+M3T3sM1+M2+M3

Answer:

The answer is the option (b) Let the equilibrium temperature of the system=T
Let T1T2<T<T3
As there is no loss to the surroundings.
Heat lost by M3=Heat gain by M1+Heat gain by M2
M3s(T3T)=M1s(TT1)+M2s(TT2)
M3sT3M3sT=M1sTM1sT1+M2sTM2sT2
T(M3+M1+M2)=M3T3+M1T1+M2T2
T=M1T1+M2T2+M3T3M1+M2+M3
Hence (b) is correct.

NCERT Exemplar Class 11 Physics Solutions Chapter 12: MCQII

Question:7

Which of the processes described below are irreversible?
(a) The increase in temperature of an iron rod by hammering it.
(b) A gas in a small container at a temperature T1 is brought in contact with a big reservoir at a higher temperature T2 which increases the temperature of the gas.
(c) A quasi-static isothermal expansion of an ideal gas in a cylinder fitted with a frictionless piston.
(d) An ideal gas is enclosed in a piston-cylinder arrangement with adiabatic walls. A weight W is added to the piston, resulting in compression of the gas.

Answer:

The answer is the option (a), (b), (d)
(a) During hammering the work done on the rod in hammering converts into heat, raises the temperature of the rod but this heat energy cannot be converted into work, so the process is not reversible.
(b) Heat flows from higher temperature to lower. When the containers are brought in contact, the heat of the bigger one gets transferred to a smaller container until the temperature in both the containers become equal, which is an average of both. But the heat cannot flow from smaller to larger.
(d) Adding weight to the piston increases the pressure and decreases the volume. It cannot be reversed back itself.

Question:8

An ideal gas undergoes isothermal process from some initial state I to final state f. Choose the correct alternatives
a) dU = 0
b) dQ = 0
c) dQ = dU
d) dQ = dW

Answer:

The answer is the option (a) and (d).
Since the process is isothermal ΔT=0or T is constant
For an ideal gas, dU=Change in internal energy=nCvdT
dT=0 ;Thus dU=0
dQ=dU+dW
dQ=dW
Hence, a and d are correct.

Question:9

Figure shows the P-V diagram of an ideal gas undergoing a change of state from A to B. Four different parts I, II, III and IV as shown in the figure may lead to the same change of state.
P - V graph for ideal gas
(a) Change in internal energy is same in IV and III cases, but not in I and II.
(b) Change in internal energy is same in all the four cases.
(c) Work done is maximum in case I
(d) Work done is minimum in case II.

Answer:

The answer is the option (b) and (c). dU is independent of the path followed in a P-V diagram. It depends only on the initial and final position. Work done is the area enclosed with the V-axis.
The initial and final position is the same for different parts I, II, III, IV. So change in U is same. Hence b is correct.
As area enclosed by the path I is maximum with V-axis, so work done during path I is maximum and minimum in III.
Hence, c is correct.

Question:10

Consider a cycle followed by an engine
P - V graph for cyclic process
1 to 2 is isothermal
2 to 3 is adiabatic
3 to 1 is adiabatic
Such a process does not exist because
a) heat is completely converted to mechanical energy in such a process, which is not possible
b) mechanical energy is completely converted to heat in this process, which is not possible
c) curves representing two adiabatic processes don’t intersect
d) curves representing an adiabatic process and an isothermal process don’t intersect

Answer:

The answer is the option (a) and (c).
The given process is cyclic, which starts at 1 and ends at 1.
So, dU=0
And dQ= dU+dW
dQ=dW
The heat energy supplied to the system gets converted to mechanical work which isn't possible by the second law of thermodynamics.

Question:11

Consider a heat engine as shown on the figure. Q1 and Q2 are heat added to heat bath T1 and heat taken from T2 in one cycle of the engine. W is the mechanical work done on the engine.
Heat Engine
If W > 0, then possibilities are:
a) Q1>Q2>0
b) Q2>Q1>0
c) Q2<Q1<0
d) Q1<0,Q2>0

Answer:

The answer is the option (a) and (c)
Q1=W+Q2 (From above figure)
W>0Q1Q2>0Q1>0
Thus, Q1>0 if both Q1 and Q2 are positive.
And Q1<Q2<0 if both are negative.
Hence, a and c are correct.

NCERT Exemplar Class 11 Physics Solutions Chapter 12: Very Short Answer

Question:12

Can a system be heated, and its temperature remains constant?

Answer:

A system can be heated, but with a constant temperature when it does work against the surrounding to compensate for the heat supplied.
ΔT=0ΔU=0
ΔQ=ΔU+ΔW
ΔQ=ΔW

Question:13

A system goes from P to Q by two different paths in the P-V diagram as shown in Figure. Heat given to the system in path 1 is 1000 J. The work done by the system along path 1 is more than path 2 by 100 J. What is the heat exchanged by the system in path 2?

P - V graph

Answer:

For path 1 Q1=1000J
WD=W1W2=100
W1=WD through path 1
W2=WD through path 2
W2=W1100 (Change in internal energy by both paths are same)
ΔU=Q1W1=Q2W21000W1=Q2(W1100)
Q2=900J

Question:14

If a refrigerator’s door is kept open, will the room become cool or hot? Explain.

Answer:

When the refrigerator door is kept open, the amount of heat absorbed from inside the refrigerator and the work is done on it by electricity both will be rejected by the refrigerator in the room. This will make the room hotter.

Question:15

Is it possible to increase the temperature of a gas without adding heat to it? Explain.

Answer:

In adiabatic compression, the temperature of gas increases while no heat is added to the system.
dQ=0
dQ=dU+dW
dU=-dW
So in compression work done on the system is negative. Thus, dU is positive, increasing the temperature of the system. As the internal energy of gas increases, its temperature decreases.

Question:16

Air pressure in a car tyre increases during driving. Explain.

Answer:

During driving, a reaction force is applied due to the force on tyres. The temperature of gas increases due to this reaction force causing the gas inside the tyre to expand while the volume inside the tyre remains constant. In accordance with the Charles Law, the temperature of the car tyre increases as well as the pressure (PT)

NCERT Exemplar Class 11 Physics Solutions Chapter 12: Short Answer

Question:17

Consider a Carnot’s cycle operating between T1=500K and T2=300K producing 1 k J of mechanical work per cycle. Find the heat transferred to the engine by the reservoirs.

Answer:

The efficiency of Carnot's engine μ=1T2/T1
T1(Temp of source)=500K
T2Temp of sink=300K
Output workInput work=1300500
1000Jx=10.6
1000x=0.4
x=10000.4=2500J

Question:18

A person of mass 60 kg wants to lose 5kg by going up and down a 10m high stairs. Assume he burns twice as much fat while going up than coming down. If 1 kg of fat is burnt on expending 7000 kilo calories, how many times must he go up and down to reduce his weight by 5 kg?

Answer:

The energy produced by 1 kg fat=7000 k cal
Energy produced by 5 kg fat=35000 k cal=35×106 cal
Energy consumed to go up one time=mgh
Energy consumed to go up and down one time=mgh+0.5mgh
E=32mgh=32×60×10×10=9000J=90004.2cal
Let us assume that he goes up and down n times, then
n×90004.2=35×106
n=35×106×4.29000=16.3×103

Question:19

Consider a cycle tyre being filled with air by a pump. Let V be the volume of the tyre (fixed) and at each stroke of the pump ΔVV)
of air is transferred to the tube adiabatically. What is the work done when the pressure in the tube is increased from P1 to P2?

Answer:

Air is transferred into tyre adiabatically.
Let the initial volume of air be V and after pumping it becomes V+dV and pressure P+dP
P1V1γ=P2V2γ
P(V+dV)γ=(P+dP)Vγ
PVγ[1+dVV]γ=P[1+dPP]Vγ
PVγ[1+γdVV]=PVγ[1+dPP]
On expanding by binomial theorem and neglecting the higher terms
1+γdVV=1+dPP
dV=VdPγP
pdV=P1P2VdPγ
W1W2dW=Vγ(P2P1)
W=Vγ(P2P1)

Question:20

In a refrigerator, one removes heat from a lower temperature and deposits to the surroundings at a higher temperature. In this process, mechanical work has to be done, which is provided by an electric motor. If the motor is of 1kW power, and heat is transferred from to , find the heat taken out of the refrigerator per second assuming its efficiency is 50% of a perfect engine.

Answer:

Here refrigerator acts as a Carnot's engine in reverse order with an efficiency η

η=1T2T1=1270300=110

η=0.5×0.1=0.05
Coefficient of performance β=Q2W=1ηη=10.050.05=19
Q2=19% WD by motor on Refrigerator
=19×1kW=19kJs

Question:22

The initial state of a certain gas is Pi,Vi,Ti. It undergoes expansion till its volume becomes Vf. Consider the following two cases:
a) The expansion takes place at constant temperature
b) The expansion takes place at constant pressure
Plot the P-V diagram for each case. In which of the two cases is the work done by the gas more?
P - V graph

Answer:

The expansion from Vi to Vf at a constant temperature in (a) is an isothermal expansion.i.e.
PiVi=PfVf at a constant temp.
The expansion at constant pressure in (b) is the isobaric process wherein the graph of P-V will be parallel to V axis until its volume becomes Vf.
The area enclosed by graph (a) is less than the graph (b). So, work done by the process (b) is more than (a).

NCERT Exemplar Class 11 Physics Solutions Chapter 12: Long Answer

Question:23

Consider a P-V diagram in which the path followed by one mole of perfect gas in a cylindrical container is shown in the figure.
P - V graph
a) find the work done when the gas is taken from state 1 to state 2
b) what is the ratio of temperature T1T2 if V2=2V1
c) given the internal energy for one mole of gas at temperature T is (32)RT, find the heat supplied to the gas when it is taken from state 1 to 2 with V2=2V1.

Answer:

Let P1V112=PBVB12=K
And P=KV12
Work done for process 1 to 2; WD=V1V2PdV=V1V2KV12dV=K[V1212]=2K[V2V1]
WD from V1to V2=
dW=2P1V112V2V1=2P2V212V2V1
  1. Equation of ideal gas PV=nRT
T=PVnR=KVnR
T1=KV1nRand T2=KV2nR
T1T2=V1V2=12
(c )
ΔU=32RΔT=32R(T2T1)
ΔU=32RT1(21)
dW=2P1V112(V2V1)
dW=2P1V112(2V1V1)
dW=2P1V1(21)=2nRT1(21)
n=1
dW=2RT1(21)
dQ=dW+dU=2RT1(21)+32RT1(21)
=(21)RT1(2+32)

Question:24

A cycle followed by an engine is shown in the figure.
P - V graph for engine
A to B: volume constant
B to C: adiabatic
C to D: volume constant
D to A: adiabatic
VC=VD=2VA=2VB
a) in which part of the cycle heat is supplied to the engine from outside?
b) in which part of the cycle heat is being given to the surrounding by the engine?
c) what is the work done by the engine in one cycle in terms of PA,PB,VA?
d) what is the efficiency of the engine?

Answer:

(a) Heat is supplied to the engine in part AB where dV=0
dW=PdV=P(0)
dW=0
dQ=dU+dW(First law of thermodynamics)
dQ=dU
P=nRTV
V=constant
pT
(b) Heat is given out by the system in a part CD where dV=0 and pressure and temperature decreases
(c ) WD by system=BAPdV+BCPdV+CDPdV+DAPdV
BAPdV=0andCDPdV=0
For adiabatic change
PVγ=K
P=KVγ

BCPdV=ABKVγdV=K[Vγ+11γ]VBVC
BCPdV=K1γ[VC1γVB1γ]
DAPdV=K1γ[VA1γVD1γ]
DAPdV=PAVAγVA1γPDVDγVD1γ1γ
DAPdV=PAVAPDVD1γ
DAPdV=PAVAPDVD1γ
BCPdV=PCVCPBVB1γ
Total WD=[PCVCPBVB+PAVAPDVD]1γ
For Adiabatic change=PBVBγ=PCVCγ
PC=PBVBγVCγ=PB(VBVC)γ=PB(VB2VB)γ=PB2γ
PD=PA2γ
Net WD= 11γ[PBVC2γPBVB+PAVAPAVD2γ]
=11γ[PB2VB2γPBVB+PAVAPA2VA2γ]
=11γ[PBVB[21γ+1]PAVA[21γ+1]]
=11γ[PAVAPBVB][121γ]
2VB=2VAVB=VA=1153[2153+1][PB+PA]VA
=32[1(12)23][PBPA]VA

Question:25

A cycle followed by an engine is shown in the figure. Find heat exchanged by the engine, with the surroundings for each section of the cycle considering Cv=(32)R.
P - V graph for engine
AB: constant volume
BC: constant pressure
CD: adiabatic
DA: constant pressure

Answer:

Ans.

(a) For AB, dV=0
dW=PdV=P×0=0
dQ=dU+dW=dU+0
dQ=dU
dQ=nCvdT
n=1;Cv=32R
dQ=1(32R)(TBTA)
dU=dQ1=32(RTBRTA)=32(PBVBPAVA)
(b) BC
dQ2=dU+dW=CvdT+PBdV
dQ2=32R(TCTB)+PB(VCVB)
=32[PCVC]32PBVBPBVB+PBVC
VA=VBandPB=PC
dQ2=52PBVC52PBVA=52PB[VCVA]
(c ) CD, adiabatic change
dQ3=0
  1. In diagram DA
ΔP=0
Compression of gas at constant pressure takes place. Therefore, heat exchange is similar to part (b)
dQ4=52PA(VAVD)

Question:26

Consider that an ideal gas is expanding in a process given by P = f(V), which passes through a point (V0,P0). Show that the gas is absorbing heat at (P0,V0) if the slope of the curve P = f(V) is larger than the slope of the adiabat passing through (P0,V0).

Answer:

Slope of graph at
(V0,P0)=dPdV(P0,V0)
P=f(V) for adiabatic process PVγ=constant K
P=KVγ
dPdV=γPV
(dPdV)(P0,V0)=γP0V0
P=f(V)
dQ=dU+dW
dQ=nCvdT+PdV
PV=nRT
T=PVnR=VnRf(V)
dTdV=1nR[f(V)+Vf(V)]
dQdV=nCv(dTdV)+P(dVdV)=nCvnR[f(V)+V(f(V))+P]
dQdV(V=V0)=CvR[f(V0)+V0(f(V0))+f(V0)]
=f(V0)[CvR+1]+V0f(V0)CvR
CPCv=RCPCV1=RCv
γ1=RCvCv=Rγ1CvR=1γ1
dQdV(V=V0)=1r1[γP0+V0f(V0)]
dQdV(V=V0)>1andγ>1so1γ1>0
γP0+V0f(V0)>0
V0f(V0)>γP0
fV0>γP0V0

Question:27

Consider one mole of perfect gas in a cylinder of unit cross-section with a piston attached. A spring is attached to the piston and to the bottom of the cylinder. Initially the spring is unstretched and the gas is in equilibrium. A certain amount of heat Q is supplied to the gas causing an increase of volume from V0 to V1.
 A spring is attached to the piston
a) what is the initial pressure of the system?
b) what is the final pressure of the system?
c) using the first law of thermodynamics, write down the relation between Q, Pa, V, V0, and k.

Answer:

  1. Piston is considered massless and balanced by atmospheric pressure. So the initial
Pressure of system inside cylinder is Pa
  1. On supplying heat volume of gas increases from V0toV1
Increase in Volume=V1V0 =Area of base×height=A×x
V1V0=A×x
x=V1V0A
(Force exerted by spring)
F=Kx=K(V1V0)A
(Force due to spring on unit area)
F=K(V1V0)
(Final total pressure on gas)
Pf=Pa+K(V1V0)
  1. dQ=dU+dW (First law of thermodynamics)
dU=Cv(TT0)
T=final temperature of the gas
T0=initial temperature of the gas
n=1
Ty=T=PfVfR=[Pa+K(V1V0)]R
dW=Pa(V1V0)+12kx2(Increase in potential energy of spring)
dQ=dU+dW=Cv(TT0)+Pa(V1V0)+12kx2
dQ=dU+dW=Cv(TT0)+Pa(V1V0)+12k(V1V0)2.

NCERT Exemplar Class 11 Physics Solutions Chapter 12 Thermodynamics: Topics

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Important Concepts & Formulas - NCERT Exemplar Class 11 Physics Chapter 12 (Thermodynamics)

1. Thermal Equilibrium - When two bodies in contact stop exchanging heat, they are in thermal equilibrium.
2. Zeroth Law of Thermodynamics - If A is in thermal equilibrium with B, and B is in thermal equilibrium with C, then A and C are also in thermal equilibrium.
3. First Law of Thermodynamics - Energy cannot be created or destroyed, only transferred as heat or work.

  • Formula: ΔU=QW (Change in internal energy = Heat added - Work done)

4. Specific Heat Capacity - The heat required to raise the temperature of 1 kg of a substance by 1C.

  • Formula: Q=mcΔT

5. Thermodynamic Processes -

  • Isothermal ( T constant): PV= constant
  • Adiabatic (No heat exchange): PVγ= constant
  • Isochoric (Volume constant): Q=ΔU
  • Isobaric (Pressure constant): Q=ΔU+W

6. Second Law of Thermodynamics - Heat flows from hot to cold objects naturally; the efficiency of a heat engine is always less than 100%.
7. Carnot Engine - An ideal engine with maximum efficiency.

  • Efficiency: η=1TCTH (Where TC= cold reservoir temp, TH= hot reservoir temp)

NCERT Exemplar Class 11 Physics Solutions Chapter-Wise

NCERT Exemplar Class 11th Solutions

Check Class 11 Physics Chapter-wise Solutions

Also, Read NCERT Solution subject-wise -

Check NCERT Notes subject-wise -

Also, Check NCERT Books and NCERT Syllabus here

Frequently Asked Questions (FAQs)

1. Is this chapter and crucial for NEET?

Yes, this chapter is important for NEET and the medical entrance exam. The physics papers tend to have 1-2 questions from this chapter every year.

2. Who has solved these NCERT questions?

Our teams of experienced teachers who have industry and teaching knowledge for years have solved these questions keeping in mind the student’s level of understanding. 

3. Are these solutions for students in school exams?

 Yes, these class 11 physics NCERT exemplar solutions chapter 12 are designed in a way so that the students can score the maximum marks as per the marking scheme.

4. How to apply the second law of thermodynamics in real life?

The second law explains why heat moves from hot to cold, why engines aren’t 100% efficient, and why systems tend toward disorder over time.

5. How does entropy change in different thermodynamic processes?

Entropy measures disorder. It increases in natural processes (like melting ice) but remains constant in ideal reversible processes.

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Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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