Physics NCERT Exemplar Solutions Class 11th Chapter Twelve: Overview, Questions, Preparation

1. Consider a P-V diagram in which the path followed by one mole of perfect gas in a cylindrical container is shown in Fig.

(a) Find the work done when the gas is taken from state 1 to state 2.

(b) What is the ratio of temperature T₁/T₂, if V₂ = 2V₁?

(c) Given the internal energy for one mole of gas at temperature T is (3/2) RT, find the heat supplied to the gas when it is taken from state 1 to 2, with V₂ = 2V₁.

Explanation- pV^1/2= constant

P=k/ $\sqrt[]{V}$

Work done from 1 to 2

W= ${\int }_{V1}^{V2}pdV=K{\int }_{V1}^{V2}\frac{dV}{\sqrt[]{V}}=k{\left[\frac{\sqrt[]{V}}{\frac{1}{2}}\right]}_{V1}^{V2}=2K\left(\sqrt[]{V_2}-\sqrt[]{V_1}\right)$

from ideal equation = pV=nRT

T= pV/nR= $\frac{p\sqrt[]{V}\sqrt[]{V}}{nR}$

T= $\frac{K\sqrt[]{V}}{nR}$

T₁= $\frac{K\sqrt[]{V1}}{nR}$ , T₁= $\frac{K\sqrt[]{V2}}{nR}$

$\frac{T_1}{T_2}=\frac{\frac{k\sqrt[]{V_1}}{nR}}{\frac{k\sqrt[]{V_2}}{nR}}={\sqrt[]{\frac{V_1}{V_2}}}^{}$ = $\sqrt[]{\frac{V_1}{2V_1}}=\frac{1}{\sqrt[]{2}}$

U= $\frac{3}{2}RT$

$∆U=U_2-U_1=\frac{3}{2}R\left(T_1-T_2\right)$

= $\frac{3}{2}$ RT₁( $\sqrt[]{V}-1$ )

$∆W$ =2p₁V₁^1/2( $\sqrt[]{V_2}-\sqrt[]{V_2}$ )

 = 2p₁V₁^1/2(2 $\sqrt[]{V_1}-\sqrt[]{V_1}$ )

 = 2p₁V₁( $\sqrt[]{2}-1$ )= 2RT₁( $\sqrt[]{2}-1$ )

$∆Q$ = $∆U+∆W$

= $\frac{3}{2}$ RT₁( $\sqrt[]{2}-1$ )+ 2RT₁( $\sqrt[]{2}-1$ )

= $\frac{7}{2}R{T}_1(\sqrt[]{2}-1)$

2. A cycle followed by an engine (made of one mole of perfect gas in a cylinder with a piston) is shown in Fig .

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? Write your answer in term of PA, PB, VA.

(d) What is the efficiency of the engine? [γ =5/3 for the gas], (C_v= 3/2R for one mole)

Explanation-a)  for the process AB

dV=0 and dW=0

dQ=dU+dW=dU

dQ=dU= change in internal energy , so heat utilised is equal to change in internal energy.

Since p= $\frac{nR}{V}T,$ in adiabatic temperature is directly proportional to pressure. So heat is supplied to the system in process AB.

b) for the process CD volume is constant but the pressure decreases, hence temperature also decreases . so heat is also given to the surroundings.

c) W_AB= ${\int }_A^{B}pdV=0$ , W_CD= ${\int }_{Vc}^{Vd}pdV=0$

W_BC= ${\int }_{VB}^{Vc}pdV=k{\int }_{VB}^{Vc}\frac{dV}{V^Y}=\frac{k}{1-Y}\left[V^{1-Y}\right]$  

= $\frac{1}{1-Y}$ [pV]= $\frac{P_c{V}_c-P_B{V}_B}{1-Y}$

W_DA= $\frac{P_A{V}_A-P_D{V}_D}{1-Y}$

B and C lies on adiabatic curve BC

P_BV_B^Y= P_CV_C^Y

P_C = P_B( $\frac{V_B}{V_C}$ )^Y = P_B( $\frac{1}{2}$ )^Y= 2^-YP_B

Total work done by the engine in one cycle ABCDA

W= W_AB+W_BC+W_CD+W_DA= W_BC+W_DA

 = $\frac{P_C{V}_C-P_B{V}_B}{1-Y}+\mathrm{}\frac{P_A{V}_A-P_D{V}_D}{1-Y}$  

W = $\frac{1}{1-Y}\left[2^{-Y}{P}_{B}2V_B-P_B{V}_B+P_A{P}_A-2^{-Y}{P}_{B}2V_B\right]$

W = $\frac{1}{1-Y}(2^{1-Y}-1)(P_B-P_A)V_A$

W= $\frac{3}{2}[1-(\frac{1}{2})$ ^2/3]( $P_B-P_A$ )V_A

3. A cycle followed by an engine (made of one mole of an ideal gas in a cylinder with a piston) is shown in Fig.. Find heat exchanged by the engine, with the surroundings for each section of the cycle. (Cv = (3/2) R)

AB : constant volume

BC : constant pressure

CD : adiabatic

DA : constant pressure .

Explanation-a)  for process AB

Volume is constant , hence work done dW=0

dQ=dU+dW=dU+0=dU

 = nC_vdT= nC_v(T_B-T_A)

 = $\frac{3}{2}R\left(T_B-T_A\right)$

= $\frac{3}{2}\left(R{T}_B-R{T}_A\right)=\frac{3}{2}\left(P_B{V}_B-P_A{V}_A\right)$

Heat exchanged = $\frac{3}{2}\left(P_B{V}_B-P_A{V}_A\right)$

b) for process BC , p =constant

dQ= dU+dW  = $\frac{3}{2}R\left(T_C-T_B\right)+P_B(V_C-V_B)$

heat exchanged = $\frac{5}{2}{P}_B\left(V_C{-V}_A\right)$

c) for process CD , because CD is adiabatic , dQ= heat exchanged =0

d) DA involves compression of gas from V_D to V_A at constant pressure P_A

heat transferred as similar way as BC₁

hence dQ = $\frac{5}{2}$ P_A(V_A-V_D)

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

Explanation- slope of the curve = f(V) , where V is the volume

Slope of P = f(V) curve at ((P_o, V₀ )= f(V_o)

Slope of adiabatic at (P_o, V₀ )= k(-Y)V_o^-1-Y =-YP_o/V_o

Now heat absorbed in the process P= f(V)

dQ=dU+dW= nC_vdT+pdV

pV=nRT

T= pV/nR

T= $\frac{1}{nR}\left[f\left(V\right)+V{f}^{\text{'}}\left(V\right)dV\right]$

$\frac{dQ}{dV}=$ nC_{v $\frac{dT}{dV}+p\frac{dV}{dV}=n{C}_V\frac{dT}{dV}+P$}

$\frac{C_V}{R}\left[f\right(V_o+V_0{f}^{\text{'}}\left(V\right)+f\left(V\right)]$

After solving we get

${\left[\frac{dQ}{dV}\right]}_{v=vo}=$ $\left[\frac{1}{Y-1}+1\right]f\left(V_o\right)+\frac{V_o{f}^{\text{'}{V}_o}}{Y-1}$

= $\frac{Y}{Y-1}{P}_o+\frac{V_o}{Y-1}{f}^{\text{'}}\left(V_o\right)$

Heat is absorbed where dQ/dV>0  when gas expands

Hence YP_o+V_of’(V_o)>0  or f’(V_o)>(-Y $\frac{P_o}{V_o}$ )