Question

Consider one mole of helium gas enclosed in a container at initial pressure $P_1$ and volume $V_1$ . It
expands isothermally to volume $4V_1$. After this, the gas expands adiabatically and its volume
becomes $32V_1$. The work done by the gas during isothermal and adiabatic expansion processes are $W_{iso}$ and $W_{adia}$ respectively. If the ratio$\frac{W_{iso}}{W_{adia}}=f\ln 2$, then $f$ is

Answer 1

For adiabatic process
$\frac{P_1}{4}(4V_1{)}^{\frac{5}{3}}=P_2(32V_1{)}^{\frac{5}{3}}$
$P_2=\frac{P_1}{4}{\left(\frac{1}{8}\right)}^{\frac{5}{3}}=\frac{P_1}{128}$
$W_{adi}=\frac{P_1{V}_1-P_2{V}_2}{\gamma -1}-\frac{P_1{V}_1-\frac{P_1}{128}\left(32V_1\right)}{\frac{5}{3}-1}$
$W_{adi}=\frac{P_1{V}_1\frac{3}{4}}{\frac{2}{3}}=\frac{9}{8}{P}_1{V}_1$
For Isothermal process
$W_{iso}=P_1{V}_1\ln \left(\frac{4V_1}{V_1}\right)=2P_1{V}_1\ln 2$
$\frac{W_{iso}}{W_{adia}}=\frac{2P_1{V}_1\ln 2}{\frac{9}{8}{P}_1{V}_1}=\frac{16}{9}\ln 2=f\ln 2$
$f=\frac{16}{9}=1.7778\approx 1.78$