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 temerature at $A$ and $B$ are $T_A=300K$ and $T_B=500K$ and pressure at $A$ is 1 atm and volume at $A$ is $4.9L$. The volume at $C$ is
(Given : $\gamma =\frac{Cp}{C_V}=\frac{5}{3},R=8.205\times 10^{-2}Latmmo{l}^{-1}$
$K^{-1},{\left(\frac{3}{2}\right)}^{{}^{2/5}}=0.81$

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

Answer 1

Answer: (B)

For A $\to$ B, Isochoric process
$\frac{P_A}{T_A}=\frac{P_B}{T_B}$
$P_B=\frac{T_B}{T_A}{P}_A$
$P_B=\frac{500}{300}\times 1=\frac{5}{3}atm...\left(i\right)$
For B $\to$ C, Adiabatic process
$\therefore \frac{T_C^{\gamma }}{P_C^{\gamma -1}}=\frac{T_B^{\gamma }}{P_B^{\gamma -1}}$
$T_C={\left(\frac{P_C}{P_B}\right)}^{{}^{\frac{\gamma -1}{\gamma }}}\times T_B$
$={\left[\frac{1}{5/3}\right]}^{{}^{\frac{\left(5/3\right)-1}{\left(5/3\right)}}}\times 500\left(Usinig\left(i\right)\right)$
$={\left(\frac{3}{5}\right)}^{2/5}\times \mathrm{500...}\left(ii\right)$
For C $\to$ A, Isobaric process
$\therefore \frac{V_C}{T_C}=\frac{V_A}{T_A}$
$V_C=V_A\times \frac{T_C}{T_A}=4.9\times {\left(\frac{3}{5}\right)}^{2/5}\times 500\times \frac{1}{300}\left(Using\left(ii\right)\right)$
$V_c=4.9\times 0.81\times \frac{5}{3}=6.6L$