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: $\left.\left(\frac{1}{3}\right)^{0.6}=0.5, \ln 2 \simeq 0.7\right]$.

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

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

Answer 1

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

For adiabatic process $( A \rightarrow B )$
$ 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 $A \rightarrow B$
$ 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} $
$\Rightarrow W _{ AB }=-60 kJ =\Rightarrow\left| W _{ AB }\right|=60 kJ$
Work done in process B $\rightarrow C$ (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 }=84 kJ$
Work done in process $C \rightarrow A$
$W _{ CA }= P \Delta V =0 (\because \Delta V =0)$
So total work done in the process $A \rightarrow B \rightarrow C$
$ W _{ ABC }= W _{ AB }+ W _{ BC }+ W _{ CA }=-60+84+0 $
$ W _{ ABC }=24 kJ$
So correct options are (B,C,D)