Question

In the given $P-V$ diagram, a monoatomic gas $\left(\gamma =\frac{5}{3}\right)$ is first compressed adiabatically from state $A$ to state $B$. Then it expands isothermally from state $B$ to state $C$. [Given: ${\left(\frac{1}{3}\right)}^{0.6}=0.5,\ln 2\simeq 0.7]$.

Which of the following statement(s) is(are) correct?

• A. The magnitude of the total work done in the process $A\to B\to C$ is $144kJ$.
• B. The magnitude of the work done in the process $B\to C$ is $84kJ$.
• C. The magnitude of the work done in the process $A\to B$ is $60kJ$.
• D. The magnitude of the work done in the process $C\to A$ is zero.

Answer 1

Answer: (B), (C), (D)

For adiabatic process $(A\to B)$
$P_A{V}_A^{\gamma }=P_B{V}_B^{\gamma }$
$10^5\times (0.8{)}^{\frac{5}{3}}=3\times 10^5{\left(V_B\right)}^{\frac{5}{3}}$
$\Rightarrow V_B=0.8\times {\left(\frac{1}{3}\right)}^{0.6}=0.4$
Work done in process $A\to B$
$W_{AB}=\frac{P_A{V}_A-P_B{V}_B}{\gamma -1}$
$\Rightarrow W_{AB}=\frac{10^5\times 0.8-3\times 10^5\times 0.4}{\frac{5}{3}-1}$
$\Rightarrow W_{AB}=-60kJ=\Rightarrow \left|W_{AB}\right|=60kJ$
Work done in process B $\to C$ (Isothermal process)
$W_{BC}=nRT\ell n\frac{V_C}{V_B}=P_B{V}_B\ell n\frac{V_C}{V_B}$
$\Rightarrow W_{BC}=3\times 10^5\times 0.4\ell n\frac{0.8}{0.4}$
$\Rightarrow W_{BC}=84kJ$
Work done in process $C\to A$
$W_{CA}=P\Delta V=0(\because \Delta V=0)$
So total work done in the process $A\to B\to C$
$W_{ABC}=W_{AB}+W_{BC}+W_{CA}=-60+84+0$
$W_{ABC}=24kJ$
So correct options are (B,C,D)