NCERT Exemplar Class 11 Physics Solutions Chapter 12 Thermodynamics
Have you ever questioned yourself about how a steam engine works, or why refrigerators cool down food, or how a hot cup of tea cools down with time? The laws of thermodynamics can explain all these phenomena in everyday life, and this lays the foundation of NCERT Class 11 Physics Chapter 12 - Thermodynamics. This chapter gives students an insight into the interdependence between heat, work, temperature, and energy in a physical system.
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
- NCERT Exemplar Class 11 Physics Solutions Chapter 12: Important Concepts and Formulas
- Advantages of NCERT Exemplar Class 11 Physics Solutions Chapter 12: Thermodynamics
- Approach to Solve Exemplar Questions of Chapter 12:Thermodynamics
- NCERT Exemplar Class 11 Physics Solutions Chapter-Wise
- NCERT Exemplar Solutions Class 11 Subject-wise
- NCERT Solutions for Class 11 Physics Chapter-wise
NCERT Exemplar Class 11 Physics Solutions Chapter 12 Thermodynamics covers important topics such as thermal equilibrium, laws of thermodynamics, heat engines, refrigerators, internal energy, and thermodynamic processes. The concepts play a fundamental role in explaining thermal processes in the real world and are commonly featured in CBSE board examinations or other competitive examinations such as JEE. The NCERT Exemplar solutions are structured in a simple step-by-step way, thus numerical problems and theory-based questions are simple to understand and answer properly. All NCERT exemplar solutions aim at reinforcing conceptual understanding, reducing prevalent errors and building a rational method of problem-solving in thermodynamics. These questions can be practised on a regular basis in order to be more analytical and gain confidence in solving problems. NCERT Exemplar Class 11 Solutions Physics on Thermodynamics is an excellent study tool in terms of revision and preparation for exams because of its high focus on real-life applications and systematic explanations.
NCERT Exemplar Class 11 Physics Solutions Chapter 12: MCQI
NCERT Exemplar Class 11 Physics Chapter 12 Thermodynamics: MCQ I, are aimed to assess basic knowledge of the thermodynamic principles in the students by multiple-choice questions with one answer. These MCQs are able to enhance the level of conceptual clarity, accuracy and speed; thus, they are of great use in quick revision and effective preparation for exams.
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, which is the isobaric process; 1 is isochoric. Out of 3 and 2, 3 has a smaller slope, hence it is isothermal. The 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 curve 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 the arrows is anticlockwise, the 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 containers A and B. Compression of gas is involved in both 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=TLet $T_{1}T_{2}<T<T_3$
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
NCERT Exemplar Class 11 Physics Solutions Chapter 12 Thermodynamics includes MCQ II type of problems, which are based on multiple-correct and reasoning-based objective questions, which would measure better conceptual learning about the laws and processes of thermodynamics. Such questions assist the students to analyse the statements closely, ignore certain misconceptions and gain confidence in tackling higher-order MCQs in exams.
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, raising the temperature of the rod, but this heat energy cannot be converted into work, so the process is not reversible.
(b) Heat flows from a higher temperature to a lower. When the containers are brought into contact, the heat of the bigger one gets transferred to the smaller container until the temperature in both containers becomes 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 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 by the V-axis.The initial and final position is the same for different parts I, II, III, and IV. So the change in U is the same. Hence, b is correct.
As the area enclosed by the path I is maximum with the 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
Examples Class 11 Physics Chapter 12 Thermodynamics: Very Short Answer are very specific and to the point explanations of fundamental thermodynamic concepts and definitions. These will assist students to better develop quick recall and conceptual accuracy, and they will be best suited to rapid revision prior to examinations.
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 surroundings 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
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 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 the 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 a 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 the tyres. The temperature of the 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 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
NCERT Exemplar Class 11 Physics Solutions Chapter 12 Thermodynamics: Short Answer provides clear and well-structured explanations to concept-based and numerical questions. These responses make students learn important concepts of thermodynamics, how to makea proper presentation, and how to prepare to score well in the exams.
Question:17
Answer:
The efficiency of Carnot's engine $\mu =1-T_2/T_1$$T_{1}$(Temp of source)=500K
$T_{2}$Temp of sink=300K
$\frac{\text{Output work}}{\text{Input work}}=1-\frac{300}{500 }$
$\frac{1000 J}{x}=1-0.6$
$\frac{1000 }{x}=0.4$
$x=\frac{1000}{0.4}=2500J$
Question:18
Answer:
The energy produced by 1 kg fat=7000 k calEnergy produced by 5 kg fat=35000 k cal=35×106 cal
Energy consumed to go up one time=mgh
Energy consumed to go up and down one time=mgh+0.5mgh
$E=\frac{3}{2}mgh=\frac{3}{2}\times 60\times 10\times 10=9000 J=\frac{9000}{4.2}cal$
Let us assume that he goes up and down n times, then
$n\times \frac{9000}{4.2}=35\times 10^{6}$
$n=35\times 10^{6}\times \frac{4.2}{9000}=16.3\times10^3$
Question:19
Answer:
Air is transferred into the tyre adiabatically.Let the initial volume of air be V, and after pumping, it becomes V+dV and the pressure P+dP
$P_1 V_1^\gamma =P_2V_{2}^{\gamma }$
$P(V+dV)^\gamma =(P+dP)V^\gamma$
$PV^\gamma \left [1+\frac{dV}{V} \right ]^\gamma =P\left [1+\frac{dP}{P} \right ]V^\gamma$
$PV^\gamma \left [1+\gamma \frac{dV}{V} \right ]=PV^\gamma\left [ 1+\frac{dP}{P} \right ]$
On expanding by the binomial theorem and neglecting the higher terms
$1+\gamma \frac{dV}{V}=1+\frac{dP}{P}$
$dV=\frac{VdP}{\gamma P}$
$\int pdV=\int_{P_{1}}^{P_{2}}\frac{VdP}{\gamma }$
$\int_{W_{1}}^{W_{2}}dW=\frac{V}{\gamma }(P_2-P_1)$
$W=\frac{V}{\gamma }\left (P_2-P_1 \right )$
Question:20
Answer:
Here refrigerator acts as a Carnot engine in reverse order with an efficiency $\eta$$\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
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 Vi to Vf 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 the 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 that of (a).
NCERT Exemplar Class 11 Physics Solutions Chapter 12: Long Answer
NCERT Exemplar Class 11 Physics Solutions Chapter 12 Thermodynamics: Long Answer provide step by step answers to in-depth questions on the laws and processes of thermodynamics. This is because these solutions are useful in enabling students to acquire effective conceptual knowledge, logical skills, and appropriate derivation skills, which are vital in the conduct of descriptive examinations.
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
$T_{1}=\frac{K\sqrt{V_{1}}}{nR }$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=0dW=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:
(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 $D\rightarrow A$
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
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, Pa, V, V0, and k.
Answer:
- Piston is considered massless and balanced by atmospheric pressure. So the initial
Pressure of the system inside the cylinder is Pa
- 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)
$dQ=dU+dW=C_v(T-T_0)+P_a(V_1-V_0)+\frac{1}{2}kx^2$
$dQ=dU+dW=C_v(T-T_0)+P_a(V_1-V_0)+\frac{1}{2}k(V_1-V_0)^2.$
NCERT Exemplar Class 11 Physics Solutions Chapter 12: Important Concepts and Formulas
Important Concepts and Formulas of NCERT Exemplar Class 11 Physics Chapter 12 Thermodynamics assist the students in reviewing the main concepts in terms of heat, work, and energy conversion in a summarised way. It is very helpful in enhancing the conceptualisation, solving number problems effectively, and performing well in board and competitive tests.
Important Concepts
1. Thermodynamic System and Surroundings
A thermodynamic system is the part of the universe under study, while everything else is called the surroundings. Systems can be open, closed, or isolated depending on the exchange of energy and matter.
2. Thermal Equilibrium and Zeroth Law of Thermodynamics
When two systems in contact do not exchange heat, they are said to be in thermal equilibrium. The Zeroth Law helps define temperature and forms the basis of thermometry.
3. Thermodynamic Processes
A thermodynamic process describes the change in state of a system. Important processes include isothermal, adiabatic, isobaric, and isochoric processes.
4. Internal Energy
Internal energy is the total energy possessed by a system due to the random motion and interaction of its particles. It depends only on the state of the system.
5. First Law of Thermodynamics
The first law of thermodynamics represents the conservation of energy in thermodynamic processes. It relates heat supplied, work done, and change in internal energy.
6. Second Law of Thermodynamics
The second law explains the direction of natural processes and introduces the concept of entropy. It states that heat cannot flow spontaneously from a colder body to a hotter body.
7. Heat Engine and Refrigerator
A heat engine converts heat into work, while a refrigerator removes heat from a colder region and transfers it to a hotter region using external work.
Important Formulas
- First Law of Thermodynamics:
$
\Delta Q=\Delta U+W
$
- Work done in isobaric process:
$
W=P\left(V_2-V_1\right)
$
- Work done in isothermal process:
$
W=n R T \ln \left(\frac{V_2}{V_1}\right)
$
- Internal energy of an ideal gas:
$
U=\frac{f}{2} n R T
$
- Mayer's relation:
$
C_P-C_V=R
$
- Efficiency of heat engine:
$
\eta=\frac{W}{Q_H}=1-\frac{Q_C}{Q_H}
$
- Coefficient of performance of refrigerator:
$
\mathrm{COP}=\frac{Q_C}{W}
$
Advantages of NCERT Exemplar Class 11 Physics Solutions Chapter 12: Thermodynamics
The chapter of thermodynamics is a concept-intensive chapter that links physics to the real-life process of heat and energy. NCERT Exemplar Class 11 Physics Solutions Chapter 12 Thermodynamics allows the students to grasp these intangible concepts in a simple and practical manner and thus, the learning process becomes efficient as well as exam-oriented.
- The solutions give understandable explanations of the laws of thermodynamics, internal energy, and thermodynamic processes that help the students understand the concepts easily without any complication.
- The Exemplar solutions also focus on the meaning and application of every principle, as opposed to rote learning of formulas, which is vital to solving theory and numerical problems.
- There are various variables and conditions in thermodynamics. The solutions divide numericals into simple steps, minimising mistakes in calculations and enhancing their accuracy.
- Sign conventions, units, and assumptions are also well-detailed so that students are not prone to the common mistakes that arise in problems involving heat, work, and energy.
- The exemplar questions are usually conceptual and slightly advanced. These solutions make students well prepared to exams of CBSE and even for entrance tests such as JEE.
- Logical reasoning is needed in the chapter to determine the right thermodynamic process. Practice also helps in improving analytical ability and decision-making during problem-solving.
- Well-structured explanations and organised answers allow quick revision before exams, saving time while reinforcing key concepts.
Approach to Solve Exemplar Questions of Chapter 12:Thermodynamics
Conceptual knowledge is usually examined by thermodynamics problems, rather than by solving lengthy equations. By means of a clear, step-by-step strategy, the students can identify the process in question properly and use the corresponding laws and formulas. Exemplar questions are simplified and easier to solve by adhering to a systematic procedure.
- The first step is to determine whether the system is open, closed, or isolated, and the process taking place, i.e. Is it an isothermal, adiabatic, isobaric or isochoric process. This determines the formulas and assumptions.
- List all given values like heat supplied, work done, pressure, volume, and temperature. Also note what is asked to avoid unnecessary calculations.
- Apply the first law of thermodynamics by properly defining the sign of heat and work. There are numerous mistakes that are a result of wrong sign conventions.
- The work and heat relations of various processes are different. Choose formulas once the identification of the thermodynamic process is ensured.
- Ensure all quantities are in SI units before substitution. Converting units early helps avoid mistakes in the final answer.
- Avoid skipping steps. Writing each step clearly improves understanding and helps in earning full marks in descriptive questions.
- Before finalising the answer, check whether the result is physically possible, such as efficiency being less than 100% or internal energy changes matching the process.
NCERT Exemplar Class 11 Physics Solutions Chapter-Wise
NCERT Exemplar Class 11 Physics Solutions Chapter-Wise Links provide students with a structured and organised way to study physics concepts chapter by chapter. These links make it easy to access well-explained solutions for numericals, conceptual questions, and derivations as per the latest NCERT and CBSE guidelines. They help students revise efficiently, strengthen conceptual understanding, and prepare confidently for board and competitive examinations.
NCERT Exemplar Solutions Class 11 Subject-wise
NCERT Exemplar Solutions Class 11 Subject-Wise Links offer a convenient and organised way for students to access solutions for Physics, Chemistry, and Mathematics in one place. The links enable the students to study each topic in a systematic manner with correct step-by-step directions that strictly adhere to the latest NCERT syllabus. They are suitable for rapid revision, clarity of the concepts and proper preparation in exams in all science subjects in Class 11.
NCERT Solutions for Class 11 Physics Chapter-wise
NCERT Solutions for Class 11 Physics Chapter-Wise Links help students study physics in a well-organised and systematic manner. By accessing solutions chapter by chapter, learners can easily understand concepts, numericals, and derivations as prescribed by the latest NCERT syllabus. These solutions support effective revision, strengthen fundamentals, and assist students in preparing confidently for school exams and competitive examinations.
Also, Read NCERT Solution subject-wise -
- NCERT Solutions for Class 11 Maths
- NCERT Solutions for Class 11 Chemistry
- NCERT Solutions for Class 11 Biology
Check NCERT Notes subject-wise -
- NCERT Notes Class 11 Maths
- NCERT Notes for Class 11 Physics
- NCERT Notes for Class 11 Chemistry
- NCERT Notes for Class 11 Biology
Also, Check NCERT Books and NCERT Syllabus here
Frequently Asked Questions (FAQs)
Q: Is this chapter and crucial for NEET?
A:
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.
Q: Who has solved these NCERT questions?
A:
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.
Q: Are these solutions for students in school exams?
A:
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.
Q: How to apply the second law of thermodynamics in real life?
A:
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.
Q: How does entropy change in different thermodynamic processes?
A:
Entropy measures disorder. It increases in natural processes (like melting ice) but remains constant in ideal reversible processes.
Articles
Upcoming School Exams
Popular Questions
A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is
| Option 1)
|
Option 2)
|
| Option 3)
|
Option 4)
|
An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range
| Option 1)
|
Option 2)
|
| Option 3)
|
Option 4)
|
A particle is projected at 600 to the horizontal with a kinetic energy . The kinetic energy at the highest point
| Option 1)
|
Option 2)
|
| Option 3)
|
Option 4)
|
In the reaction,
| Option 1)
|
Option 2)
|
| Option 3)
|
Option 4)
|
How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?
| Option 1)
0.02 |
Option 2)
3.125 × 10-2 |
| Option 3)
1.25 × 10-2 |
Option 4)
2.5 × 10-2 |
Colleges After 12th
Courses After 12th
Applications for Admissions are open.
As per latest syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
JEE Main Important Chemistry formulas
Get nowAs per latest syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters
JEE Main high scoring chapters and topics
Get nowAs per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE
JEE Main Important Mathematics Formulas
Get nowAs per latest syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters