Question

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

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

Answer 1

Answer: (B)

For $A\to B$, 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 $B\to C$, adiabatic process
$\therefore \frac{T_C^{\gamma }}{P_C^{\gamma -1}}=\frac{T_B^{\gamma }}{P_B^{\gamma -1}}$
or $T_C={\left(\frac{P_C}{P_B}\right)}^{\frac{\gamma -1}{\gamma }}\times T_B={\left[\frac{1}{5/3}\right]}^{\frac{\left(5/3\right)-1}{\left(5/3\right)}}\times 500$ (Using (i))
$={\left(\frac{3}{5}\right)}^{2/5}\times 500\dots (ii)$
For $C\to A$, 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)
$=4.9\times 0.81\times \frac{5}{3}=6.6L$