Question

One mole of an ideal gas at $900\, K$, 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

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