Question

One mole of an ideal gas at $900K$, undergoes two reversible processes, $I$ followed by $II$, as shown below. If the work done by the gas in the two process are same, the value of $\ln \frac{V_3}{V_2}$ ____

( $U$ : internal energy, $S$ : entropy, $p$ : pressure, $V:$ volume, $R$ : gas constant)
(Given: molar heat capacity at constant volume, $C_{v,m}$ of the gas is $\frac{5}{2}R$ )

Answer 1

Answer: 10

$1^{\text{st }}$ process is adiabatic since entropy is constant.
$W_1=\Delta U$
$\Delta U=450R-2250R=-1800R$
$W_1=-1800R$ .... (1)
In $2^{\text{nd }}$ process internal energy is constant it means it is a isothermal process.
$W_2=-2.303nRT\log \frac{V_3}{V_2}$ ... (2)
$=-nRT\ln \frac{V_3}{V_2}$ ... (3)
Given, $n=1$ mole, here temperature is unknown
$U=n{C}_{V}T$ for process $II$
$450R=1\times \frac{5}{2}RT$
$T=\frac{450\times 2}{5}=180K$
Equation (1) = equation (2)
$W_1=W_2$
$-1800R=-1\times R\times 180\ln \frac{V_3}{V_2}$
$\ln \frac{V_3}{V_2}=\frac{1800}{180}=10$