Question

0.2 moles of an ideal gas is taken round the cycle $ABC$ as shown in the figure. The path $B \rightarrow C$ is an adiabatic process, $A \rightarrow B$ is an isochoric process and $C \rightarrow A$ is an isobaric process. The temperature at $A$ and $B$ are $T_{A}=300\, K$ and $T_{B}=500\, K$ and pressure at $A$ is $1 atm$ and volume at $A$ is 4.9 L. The volume at $C$ is (Given: $\gamma=\frac{C_{P}}{C_{V}}=\frac{5}{3}, R=8.205 \times 10^{-2} L \,atm\, mol ^{-1} K ^{-1}$ $\left(\frac{3}{2}\right)^{2 / 5}=0.81)$

• A. 6.9 L
• B. 6.6 L
• C. 5.5 L
• D. 5.8 L

Answer 1

Answer: (B)

For $A \rightarrow B$, volume is constant
$\therefore \frac{P_{A}}{T_{A}}=\frac{P_{B}}{T_{B}} $ or
$ P_{B}=\frac{T_{B}}{T_{A}} \times P_{A}$
$=\frac{500}{300} \times 1=\frac{5}{3} \,atm \,\,\,\,\,\,\,\,\dots(i)$
For $B \rightarrow C$, adiabatic process
$\therefore \frac{T_{C}^{\gamma}}{P_{C}^{\gamma-1}}=\frac{T_{B}^{\gamma}}{P_{B}^{\gamma-1}}$
or $T_{C}=\left(\frac{P_{C}}{P_{B}}\right)^{\frac{\gamma-1}{\gamma}} \times T_{B}=\left[\frac{1}{5 / 3}\right]^{\frac{\left(5 / 3\right)-1}{\left(5 / 3\right)} }\times500$ (Using (i))
$=\left(\frac{3}{5}\right)^{2 / 5} \times 500\,\,\,\,\,\,\,\dots(ii)$
For $C \rightarrow A$, pressure is constant
$\therefore \frac{V_{C}}{T_{C}}=\frac{V_{A}}{T_{A}} $
or $ V_{C}=V_{A} \times \frac{T_{C}}{T_{A}}$
$=4.9 \times\left(\frac{3}{5}\right)^{2 / 5} \times 500 \times \frac{1}{300} $ (Using (ii)
$=4.9 \times 0.81 \times \frac{5}{3}=6.6\, L$