NCERT Exemplar Class 11 Physics Solutions Chapter 12 Thermodynamics 2021

NCERT Exemplar Class 11 Physics Solutions Chapter 12 Thermodynamics

Edited By Safeer PP | Updated on Aug 09, 2022 11:14 AM IST

NCERT Exemplar Class 11 Physics solutions chapter 12 - Thermodynamics is one of the most sought topics in physics for both CBSE exams and entrance exams. This chapter of NCERT Class 11 Physics Syllabus explains various topics that can be turned into a question for the students. In this NCERT exemplar Class 11 Physics chapter 12 solutions, the students will learn about the relation between energy, matter, heat, and work. This entire Class 11 Physics NCERT exemplar solutions chapter 12 is based on the process of turning heat into energy and how this entire process affects matters.
Also, check NCERT Solutions for Class 11 other subjects.

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This Story also Contains

NCERT Exemplar Class 11 Physics Solutions Chapter 12 MCQI
NCERT Exemplar Class 11 Physics Solutions Chapter 12 MCQII
NCERT Exemplar Class 11 Physics Solutions Chapter 12 Very Short Answer
NCERT Exemplar Class 11 Physics Solutions Chapter 12 Short Answer
NCERT Exemplar Class 11 Physics Solutions Chapter 12 Long Answer
Topics Covered in NCERT Exemplar Class 11 Physics Solutions Chapter 12 Thermodynamics
What Will The student learn in NCERT Exemplar Class 11 Physics Solutions Chapter 12 Thermodynamics?
NCERT Exemplar Class 11 Physics Solutions Chapter-Wise
Important Topics To Cover From NCERT Exemplar Solutions for Class 11 Physics Chapter 12
NCERT Exemplar Class 11 Solutions

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.

(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 $14.5 \times 10^{3} cal/min.$ This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming 1 kg requires $580 \times 10^{3} cal$ l 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

$\text{Amount of sweat produced}=\frac{14.5\times10^3}{580\times10^3}=0.025Kg$

Question:3

Consider P-V diagram for an ideal gas shown in Figure.

Out of the following diagrams (Figure), which represents the T-P diagram?

(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

a) $6P_{0}V_{0}$
b) $-2P_{0}V_{0}$
c) $+2P_{0}V_{0}$
d) $+4P_{0}V_{0}$

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
$=-2P_{0}V_{0}$

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) $2^{\gamma-1}$
b) $\left (\frac{1}{2} \right )^{\gamma-1}$
c) $\left (\frac{1}{1-\gamma} \right )^{2}$
d) $\left (\frac{1}{\gamma-1} \right )^{2}$

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 $1\rightarrow 2$

$P_1V_1=P_{2}V_{2}$

$P_{0}(2V_{0})=P_{2}V_{0}$

$P_{2}=2P_{0}$
For adiabatic process $1\rightarrow 2$
$P_{1}V_{1}^{\gamma }=P_2V_{2}^{\gamma}$

$P_{0}(2V_{0})^{\gamma }=P_{2}(V_{0})^{\gamma }$

$P_{2}=\left (\frac{2V_{0}}{V_{0}} \right )^{\gamma} P_{0}=2^{\gamma } P_{0}$

$\frac{(P_{2})_{B}}{(P_{1})_{A}}=\frac{2^{\gamma } P_{0}}{2P_{0}}=2^{\gamma -1}$

Question:6

Three copper blocks of masses $M_{1}, M_{2} and M_{3}$ kg respectively are brought into thermal contact till they reach equilibrium. Before contact, they were at $T_{1}, T_{2}, T_{3} (T_{1} > T_{2} > T_{3} )$ . Assuming there is no heat loss to the surroundings, the equilibrium temperature T is (s is specific heat of copper)

a) $T=\frac{T_{1}+T_{2}+ T_{3}}{3}$
b) $T=\frac{M_{1}T_{1}+M_{2}T_{2}+ M_{3}T_{3}}{M_{1}+M_{2}+M_{3}}$
c) $T=\frac{M_{1}T_{1}+M_{2}T_{2}+ M_{3}T_{3}}{3(M_{1}+M_{2}+M_{3})}$
d) $T=\frac{M_{1}T_{1}s+M_{2}T_{2}s+ M_{3}T_{3}s}{M_{1}+M_{2}+M_{3}}$

Answer:

The answer is the option (b) Let the equilibrium temperature of the system=T
Let $T_{1}T_{2}<T<T_3$
As there is no loss to the surroundings.
Heat lost by $M_3$ =Heat gain by $M_{1}$ +Heat gain by $M_{2}$
$M_{3}s(T_{3}-T)=M_{1}s(T-T_{1})+M_{2}s(T-T_{2})$
$M_3sT_3-M_3sT=M_1sT-M_1sT_1+M_2sT-M_2sT_2$
$T(M_{3}+M_{1}+M_2)=M_3T_3+M_1T_1+M_2T_2$
$T=\frac{M_1T_1+M_2T_2+M_{3}T_{3}}{M_{1}+M_2+M_3}$
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 $\Delta T=0$ or 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.

(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

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. $Q_{1}$ and $Q_{2}$ are heat added to heat bath $T_{1}$ and heat taken from $T_{2}$ in one cycle of the engine. W is the mechanical work done on the engine.

If W > 0, then possibilities are:
a) $Q_{1} > Q_{2} > 0$
b) $Q_{2} > Q_{1} > 0$
c) $Q_{2} < Q_{1} < 0$
d) $Q_{1} < 0, Q_{2} > 0$

Answer:

The answer is the option (a) and (c)
$Q_{1}=W+Q_{2}$ (From above figure)
$W>0\\ Q_{1}-Q_{2}>0\\ Q_{1}>0$
Thus, $Q_{1}>0$ if both $Q_{1}$ and $Q_{2}$ are positive.
And $Q_{1}<Q_{2}<0$ 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.
$\Delta T=0 \Rightarrow \Delta U=0$
$\Delta Q=\Delta U+\Delta W$
$\Delta Q=\Delta W$

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?

Answer:

For path 1 Q1=1000J
$WD=W_1-W_2=100$
$W_1=WD$ through path 1
$W_2=WD$ through path 2
$W_{2}=W_{1}-100$ (Change in internal energy by both paths are same)
$\Delta U=Q_{1}-W_{1}=Q_{2}-W_{2}\\ 1000-W_{1}=Q_{2}-(W_1-100)$
$Q_2=900J$

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 $(P\propto T)$

NCERT Exemplar Class 11 Physics Solutions Chapter 12 Short Answer

Question:17

Consider a Carnot’s cycle operating between $T_1 = 500K$ and $T_{2} = 300K$ 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 $\mu =1-T_2/T_1$
$T_{1}$ (Temp of source)=500K
$T_{2}$ Temp of sink=300K
$\frac{\text{Output work}}{\text{Input work}}=1-\frac{300}{500 }$
$\frac{1000 J}{x}=1-0.6$
$\frac{1000 }{x}=0.4$
$x=\frac{1000}{0.4}=2500J$

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
$E=\frac{3}{2}mgh=\frac{3}{2}\times 60\times 10\times 10=9000 J=\frac{9000}{4.2}cal$
Let us assume that he goes up and down n times, then
$n\times \frac{9000}{4.2}=35\times 10^{6}$
$n=35\times 10^{6}\times \frac{4.2}{9000}=16.3\times10^3$

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 $\Delta V\leq V$ )
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
$P_1 V_1^\gamma =P_2V_{2}^{\gamma }$
$P(V+dV)^\gamma =(P+dP)V^\gamma$
$PV^\gamma \left [1+\frac{dV}{V} \right ]^\gamma =P\left [1+\frac{dP}{P} \right ]V^\gamma$
$PV^\gamma \left [1+\gamma \frac{dV}{V} \right ]=PV^\gamma\left [ 1+\frac{dP}{P} \right ]$
On expanding by binomial theorem and neglecting the higher terms
$1+\gamma \frac{dV}{V}=1+\frac{dP}{P}$
$dV=\frac{VdP}{\gamma P}$
$\int pdV=\int_{P_{1}}^{P_{2}}\frac{VdP}{\gamma }$
$\int_{W_{1}}^{W_{2}}dW=\frac{V}{\gamma }(P_2-P_1)$
$W=\frac{V}{\gamma }\left (P_2-P_1 \right )$

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 $\eta$

$\eta =1-\frac{T_2}{T_1}=1-\frac{270}{300}=110$

$\eta '=0.5\times0.1=0.05$
Coefficient of performance $\beta =\frac{Q_2}{W}=\frac{1-\eta '}{\eta '}=\frac{1-0.05}{0.05}=19$
Q₂=19% WD by motor on Refrigerator
$=19\times 1kW=19kJs$

Question:21

If the coefficient of performance of a refrigerator is 5 and operates at the room temperature, find the temperature inside the refrigerator.

Answer:

$\beta=\frac{T_2}{T_1-T_2}$
$T_1=27+273=300K$
$5=\frac{T_{2}}{300-T_{2}}$
On solving, we get
$T_2=250K$

Question:22

The initial state of a certain gas is $P_i, V_i, T_i.$ It undergoes expansion till its volume becomes $V_f.$ 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?

Answer:

The expansion from V_ito V_f at a constant temperature in (a) is an isothermal expansion.i.e.
$P_{i}V_{i}=P_fV_f$ 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 $V_f.$
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.

a) find the work done when the gas is taken from state 1 to state 2
b) what is the ratio of temperature $\frac{T_{1}}{T_{2}}$ if $V_{2}=2V_{1}$
c) given the internal energy for one mole of gas at temperature T is $\left (\frac{3}{2} \right )RT$ , find the heat supplied to the gas when it is taken from state 1 to 2 with $V_2 = 2V_1$ .

Answer:

Let $P_{1}V_{1}^{\frac{1}{2}}=P_BV_B^{\frac{1}{2}}=K$
And $P=\frac{K}{V^{\frac{1}{2}}}$
Work done for process 1 to 2; $WD=\int_{V_{1}}^{V_{2}}PdV=\int_{V_{1}}^{V_{2}}\frac{K}{V^{\frac{1}{2}}}dV=K\left [ \frac{V^{\frac{1}{2}}}{\frac{1}{2}} \right ]=2K\left [ \sqrt{V_{2}}-\sqrt{V_{1}} \right ]$
WD from V₁to V₂=
$dW=2P_1V_1^{\frac{1}{2}}V_2-V_1=2P_2V_2^{\frac{1}{2}}V_2-V_1$

Equation of ideal gas PV=nRT

$T=\frac{PV}{nR}=\frac{K\sqrt{V}}{nR }$
$T_{1}=\frac{K\sqrt{V_{1}}}{nR }$ and $T_{2}=\frac{K\sqrt{V_{2}}}{nR }$
$\frac{T_1}{T_2}=\frac{\sqrt{V_1}}{\sqrt{V_2}}=\frac{1}{\sqrt{2}}$
(c )
$\Delta U=\frac{3}{2}R\Delta T=\frac{3}{2}R(T_{2}-T_{1})$
$\Delta U=\frac{3}{2}RT_1(\sqrt{2}-1)$
$dW=2P_1V_1\frac{1}{2}\left (\sqrt{V_2}-\sqrt{V_1} \right )$
$dW=2P_1V_1^{\frac{1}{2}}(\sqrt{2V_1}-\sqrt{V_1})$
$dW=2P_1V_1(\sqrt{2}-1) =2nRT_{1}(\sqrt{2}-1)$
$n=1$
$dW=2RT_1(\sqrt{2}-1)$
$dQ=dW+dU=2RT_{1}(\sqrt{2}-1)+\frac{3}{2}RT_{1}\left (\sqrt{2}-1 \right )$
$=(\sqrt{2}-1)RT_{1}\left (2+\frac{3}{2} \right )$

Question:24

A cycle followed by an engine is shown in the figure.

A to B: volume constant
B to C: adiabatic
C to D: volume constant
D to A: adiabatic
$V_C = V_D = 2V_A = 2V_B$
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 $P_A, P_B, V_A$ ?
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
$P=\frac{nRT}{V}$
V=constant
$p\propto T$
(b) Heat is given out by the system in a part CD where dV=0 and pressure and temperature decreases
(c ) WD by system= $\int_{B}^{A}PdV+\int_{B}^{C}PdV+\int_{C}^D{}PdV+\int_{D}^{A}PdV$
$\int_{B}^{A}PdV=0 \text{and}\int_{C}^D{}PdV=0$
For adiabatic change
$PV^{\gamma}=K$
$P=\frac{K}{V^{\gamma}}$

$\int_{B}^{C}PdV=\int_{A}^{B}\frac{K}{V^{\gamma }}dV=K\left [ \frac{V^{-\gamma +1}}{1-\gamma } \right ]^{V_{C}}_{V_{B}}$
$\int_{B}^{C}PdV=\frac{K}{1-\gamma}\left [ V_{C}^{1-\gamma}-V_{B}^{1-\gamma} \right ]$
$\int_{D}^{A}PdV=\frac{K}{1-\gamma}\left [ V_{A}^{1-\gamma}-V_{D}^{1-\gamma} \right ]$
$\int_{D}^{A}PdV=\frac{P_{A}V_{A}^{\gamma}V_{A}^{1-\gamma}-P_{D}V_{D}^{\gamma}V_{D}^{1-\gamma}}{1-\gamma}$
$\int_{D}^{A}PdV=\frac{P_{A}V_{A}-P_{D}V_{D}}{1-\gamma}$
$\int_{D}^{A}PdV=\frac{P_{A}V_{A}-P_{D}V_{D}}{1-\gamma}$
$\int_{B}^{C}PdV=\frac{P_{C}V_{C}-P_{B}V_{B}}{1-\gamma}$
Total $WD=\frac{[P_{C}V_{C}-P_{B}V_{B}+P_{A}V_{A}-P_{D}V_{D}]}{1-\gamma}$
For Adiabatic change= $P_{B}V_{B}^\gamma =P_{C}V_{C}^{\gamma}$
$P_C=\frac{P_BV_B^{\gamma}}{V_C^\gamma}=P_B\left (\frac{V_B}{V_C} \right )^\gamma=P_B\left (\frac{V_B}{2V_B} \right )^\gamma=\frac{P_B}{2^\gamma}$
$P_{D}=\frac{P_{A}}{2^\gamma }$
Net WD= $\frac{1}{1-\gamma} [P_BV_C2-\gamma-P_BV_B+P_AV_A-P_AV_D2-\gamma ]$
$=\frac{1}{1-\gamma}[P_B2V_B2^{-\gamma}-P_BV_B+P_AV_A-P_A2V_A2^{-\gamma}]$
$=\frac{1}{1-\gamma}[-P_{B}V_{B}[-2^{1-\gamma }+1]P_{A}V_{A}[-2^{1-\gamma }+1]]$
$=\frac{1}{1-\gamma}[P_{A}V_{A}-P_{B}V_{B}][1-2^{1-\gamma}]$
$2V_B=2V_A\Rightarrow V_B=V_A=\frac{1}{1-\frac{5}{3}}\left [ -2^{1-\frac{5}{3}}+1 \right ][-P_B+P_A]V_A$
$=\frac{3}{2}\left [ 1-\left ( \frac{1}{2} \right )^{\frac{2}{3}} \right ][P_B-P_A]V_A$

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 $C_{v} = \left (\frac{3}{2} \right )R$ .

AB: constant volume
BC: constant pressure
CD: adiabatic
DA: constant pressure

Answer:

Ans.

(a) For $A\rightarrow B$ , dV=0
$dW=\int PdV=\int P\times 0=0$
$dQ=dU+dW=dU+0$
dQ=dU
$dQ=nC_vdT$
$n=1;C_v=\frac{3}{2}R$
$dQ=1\left (\frac{3}{2} R \right )(T_B-T_A)$
$dU=dQ_1=\frac{3}{2}\left (RT_B-RT_A \right )=\frac{3}{2}(P_BV_B-P_AV_A)$
(b) $B\rightarrow C$
$dQ_2=dU+dW=C_vdT+P_BdV$
$dQ_2=\frac{3}{2}R\left (T_C-T_B \right )+P_B\left (V_C-V_B \right )$
$=\frac{3}{2}[P_CV_C]-\frac{3}{2}{P_BV_B}-P_BV_B+P_BV_C$
$V_A=V_B and P_B=P_{C}$
$dQ_2=\frac{5}{2}P_BV_C-\frac{5}{2}P_BV_A=\frac{5}{2}P_B[V_C-V_A ]$
(c ) $C\rightarrow D$ , adiabatic change
$dQ_{3}=0$

In diagram $D\rightarrow A$

$\Delta P=0$
Compression of gas at constant pressure takes place. Therefore, heat exchange is similar to part (b)
$dQ_{4}=\frac{5}{2}P_A(V_A-V_D)$

Question:26

Consider that an ideal gas is expanding in a process given by P = f(V), which passes through a point $(V_0, P_0)$ . Show that the gas is absorbing heat at $(P_0, V_0)$ if the slope of the curve P = f(V) is larger than the slope of the adiabat passing through $(P_0, V_0)$ .

Answer:

Slope of graph at
$(V_0,P_0)=\frac{dP}{dV}_{(P_0,V_0)}$
$P=f(V)$ for adiabatic process $PV^{\gamma }=\text{constant }K$
$P=\frac{K}{V^{\gamma}}$
$\frac{dP}{dV}=-\frac{\gamma P}{}V$
$(\frac{dP}{dV})_{(P_0,V_0)}=-\gamma \frac{P_0}{V_0 }$
$P=f(V)$
$dQ=dU+dW$
$dQ=nC_vdT+PdV$
$PV=nRT$
$T=\frac{PV}{nR}=\frac{V}{nR}f(V)$
$\frac{dT}{dV}=\frac{1}{nR}[f(V)+Vf'(V)]$
$\frac{dQ}{dV}=nC_v\left (\frac{dT}{dV} \right )+P\left (\frac{dV}{dV} \right )=\frac{nC_v}{nR}[f(V)+V(f'(V))+P]$
$\frac{dQ}{dV}_{(V=V_0)}=\frac{C_v}{}R\left [f(V_0)+V_0(f'(V_0))+f(V_0) \right ]$
$=f(V_0)\left [\frac{C_v}{}R+1 \right ]+V_0f'(V_0)\frac{C_v}{}R$
$C_P-C_v=R\Rightarrow \frac{C_P}{C_V}-1=\frac{R}{C_v}$
$\gamma -1=\frac{R}{C_v}\Rightarrow C_v=\frac{R}{\gamma -1}\Rightarrow \frac{C_v}{}R=\frac{1}{\gamma -1}$
$\frac{dQ}{dV}_{(V=V_0)}=\frac{1}{r-1}\left [\gamma P_0+V_0f'{(V_0)} \right ]$
$\frac{dQ}{dV}_{(V=V_0)}>1 and \gamma >1 so\frac{1}{\gamma -1}>0$
$\gamma P_0+V_0f'(V_0)>0$
$V_0f'(V_0)>-\gamma P_0$
$f'V_0>-\frac{\gamma P_0}{V_0}$

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) 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, P_a, V, V₀, and k.

Answer:

Piston is considered massless and balanced by atmospheric pressure. So the initial

pressure of system inside cylinder is P_a

On supplying heat volume of gas increases from $V_{0}to V_1$

Increase in Volume= $V_1-V_{0}$ =Area of base×height= $A\times x$
$V_{1}-V_{0}=A\times x$
$x=\frac{V_{1}-V_{0}}{A}$
(Force exerted by spring)
$F=Kx=\frac{K(V_1-V_0)}{A}$
(Force due to spring on unit area)
$F=K(V_{1}-V_0)$
(Final total pressure on gas)
$P_f=P_a+K(V_1-V_0)$

$dQ=dU+dW$ (First law of thermodynamics)

$dU=C_v(T-T_0)$
T=final temperature of the gas
T₀=initial temperature of the gas
n=1
$T_y=T=\frac{P_fV_f}{R}=\frac{\left [P_a+K(V_1-V_0) \right ]}{R}$
$dW=P_a(V_1-V_0)+\frac{1}{2}kx^{2}$ (Increase in potential energy of spring)
$dQ=dU+dW=C_v(T-T_0)+P_a(V_1-V_0)+\frac{1}{2}kx^2$
$dQ=dU+dW=C_v(T-T_0)+P_a(V_1-V_0)+\frac{1}{2}k(V_1-V_0)^2.$

One of the most crucial topics covered is thermodynamic equilibrium, which explains when a system is isolated from the surrounding, the properties do not change. Other than this, various other topics are covered like entropy, enthalpy, laws of thermodynamics, thermodynamic processes, types of system, etc. are covered in Class 11 Physics NCERT exemplar solutions chapter 12.

We have solved all the questions that the NCERT book has after the chapter. These questions play a major role in understanding the chapter in detail, and also in answering questions in exams. Students can easily check their answers against the ones that our experts have solved. Also, one can download the NCERT exemplar Class 11 Physics solutions chapter 12 pdf download for offline access using the download link.

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Topics Covered in NCERT Exemplar Class 11 Physics Solutions Chapter 12 Thermodynamics

1. Introduction
2. Thermal equilibrium
3. Zeroth law of thermodynamics
4. Heat, internal energy, and work
5. The first law of thermodynamics
6. Specific heat capacity
7. Thermodynamic state variables and equation of state
8. Thermodynamic processes
9. Heat engines
10. Refrigerators and heat pumps
11. The second law of thermodynamics
12. Reversible and irreversible processes
13. Carnot engine

What Will The student learn in NCERT Exemplar Class 11 Physics Solutions Chapter 12 Thermodynamics?

In this chapter, several topics revolve around heat and energy and interconversion between them. In thermodynamics, the rate of the change or how the change happens is not accounted for. Instead, it takes into account the final and initial stage of a process in bulk and not at microscopic levels.
In NCERT exemplar Class 11 Physics chapter 12 solutions, the learners will delve into various thermodynamic systems like closed system, open systems, and isolated systems.
The thermodynamic process like isothermal, isobaric, isochoric, and adiabatic process is explained in detail in Class 11 Physics NCERT exemplar solutions chapter 12.
Two of the major topics are explained, that is enthalpy and entropy. Enthalpy is the measurement of energy, and entropy is the measurement of disorder in a system. Other than these factors, 4 laws of thermodynamics and state of equilibriums are topics that hold the most importance.
Also, in the NCERT exemplar Class 11 Physics solutions chapter 12, various applications of thermodynamics are added like heat engines, refrigerators, and Carnot engines.

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NCERT Exemplar Class 11 Physics Solutions Chapter-Wise

Chapter 2 Units and Measurement

Chapter 3 Motion in a Straight Line

Chapter 4 Motion in a Plane

Chapter 5 Laws of Motion

Chapter 6 Work, Energy, and Power

Chapter 7 Systems of Particles and Rotational Motion

Chapter 8 Gravitation

Chapter 9 Mechanical Properties of Solids

Chapter 10 Mechanical Properties of Fluids

Chapter 11 Thermal Properties of Matter

Chapter 13 Kinetic Theory

Chapter 14 Oscillations

Chapter 15 Waves

Important Topics To Cover From NCERT Exemplar Solutions for Class 11 Physics Chapter 12

· One of the very things that students should learn from this chapter and NCERT exemplar Class 11 Physics solutions chapter 12 is the factors, terminology, processes, and variables. Enthalpy, entropy, specific heat, thermodynamic processes, etc., are some topics that will help in higher education and a better understanding of the chapter.

· Secondly, students should focus on all the laws of thermodynamics. There are in total four laws which hold importance for exams, entrance exams, and from an application point of view in the field of engineering and science.

· Lastly, the students should focus on the real-world applications of thermodynamics with the help of NCERT exemplar Class 11 Physics chapter 12 solutions. One should learn in detail about the heat engines, refrigerators, and the Carnot engine. Focus on the working process, the importance, and its application.

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Check Class 11 Physics Chapter-wise Solutions

Chapter 1	Physical world
Chapter 2	Units and Measurement
Chapter 3	Motion in a straight line
Chapter 4	Motion in a Plane
Chapter 5	Laws of Motion
Chapter 6	Work, Energy and Power
Chapter 7	System of Particles and Rotational motion
Chapter 8	Gravitation
Chapter 9	Mechanical Properties of Solids
Chapter 10	Mechanical Properties of Fluids
Chapter 11	Thermal Properties of Matter
Chapter 12	Thermodynamics
Chapter 13	Kinetic Theory
Chapter 14	Oscillations
Chapter 15	Waves

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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. What all questions are solved in solutions?

All 27 questions given in the exercise of the physics NCERT exemplar book is solved with utmost detail. All the NCERT exemplar solutions for class 11 physics, chapter 12 are in simple language and easy to understand.

3. 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.

4. 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.

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NCERT Exemplar Class 11 Physics Solutions Chapter 12 Thermodynamics

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Topics Covered in NCERT Exemplar Class 11 Physics Solutions Chapter 12 Thermodynamics

What Will The student learn in NCERT Exemplar Class 11 Physics Solutions Chapter 12 Thermodynamics?

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Important Topics To Cover From NCERT Exemplar Solutions for Class 11 Physics Chapter 12

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