Physics NCERT Exemplar Solutions Class 11th Chapter Twelve

This is a short answer type question as classified in NCERT Exemplar

For path1

Heat Q₁= 1000J

Work done =W₁

For path 2

Work done W₂= W₁-100

As change in internal energy is same

dU=Q₁-W₁=Q₂-W₂

1000-W₁=Q₂-W₁+100

Q₂= 1000-100= 900J

This is a short answer type question as classified in NCERT Exemplar

Yes this is possible when the entire heat supplied to the system is utilised in expansion.

So its working against the surroundings.

This is a long answer type question as classified in NCERT Exemplar

(a) Initially the piston is in equilibrium P_i=P_a

(b) On supplying heat , the gas expands from V_o to V_i

so increase in volume of the gas =V_i-V_o

as the piston is of unit cross sectional area hence extension in the spring

x= $\frac{V_i-V_0}{area}=V_i-V_0$

force exerted by the spring on the piston= F= kx= K(V_i - V_o)

hence final pressure =P_f =P_a +kx

= P_a+K $*$ ( $V_i-V_0$ )

(c) From first law of thermodynamics

dQ=dU+dW

dU=C_v(T-T_o) = C_v(T-T_o)

T= $\frac{P_f{V}_1}{R}=\left[\frac{P_a+K\left(V_i-V_0\right)}{R}\right]\frac{V_1}{R}$

Work done by the gas =pdV+ increase in PE of the spring

= P_a(V₁-V_o) + $\frac{1}{2}k$ x²

dQ=dU+dW

= C_v(T-T_o)+P_a(V-V_o)+ $\frac{1}{2}k$ x²

= C_v(T-T_o)+P_a(v-V_o)+1/2 ( $V_i-V_0$ )²

This is a long answer type question as classified in NCERT Exemplar

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 $\frac{P_o}{V_o}$ )

This is a long answer type question as classified in NCERT Exemplar

(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)

This is a long answer type question as classified in NCERT Exemplar

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

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