In the given P-V diagram, a monoatomic gas ( =5 3 ) is first comp… - Vidyadip

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

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

$(A)$ The magnitude of the total work done in the process $A \rightarrow B \rightarrow C$ 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

For adiabaric process $( A \rightarrow B )$

$P_A V_x^t=P_x V_x^t$

$10^2 \times(0.8)^{\frac{2}{2}}=3 \times 10^2\left(V_n\right)^{\frac{2}{1}}$

$\Rightarrow V_x=0.8 \times\left(\frac{1}{3}\right)^{0.2}=0.4$

Wodk doue in process $A \rightarrow B$

$W_{c x}=\frac{P_s V_0-P_x V_x}{\gamma-1}$

$\Rightarrow W_{\text {sx }}=\frac{10^2 \times 0.8-3 \times 10^2 \times 0.4}{\frac{5}{3}-1}$

$\Rightarrow W_{\text {As }}=-60 lJ =\Rightarrow\left|W_{\lambda \Omega}\right|=60 lJ$

Work done in process $B \rightarrow C$ (Isothermal process)

$W_{x=}=n R T / n \frac{V_8}{V_x}=P_x V_x \ell m \frac{V_8}{V_x}$

$\Rightarrow W_{x c}=3 \times 10^2 \times 0.4 \ln \frac{0.8}{0.4}$

$\Rightarrow W_{s c}=34 kJ$

Wodk doue in process $C \rightarrow A$

$W_{C_A}=P \Delta V=0 \quad(\because \Delta V=0)$

So toral work done in the process $A \rightarrow B \rightarrow C$

$W_{A B C}=W_{A \triangle}+W_{y C}+W_{C A}=-60+84+0$

$W_{A B C}=24 kJ$

So comect options are $(B,C.D)$

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