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 $4 V_{1}$. After this, the gas expands adiabatically and its volume becomes $32 V _{1}$. The work done by the gas during isothermal and adiabatic expansion processes are $W_{\text {iso }}$ and $W$ respectively. If the ratio $\frac{W_{\text {iso }}}{W_{\text {adia }}}=f \ln 2$, then $f$ is ?

Answer 1

$W_{\text {iso }}=R T \ln 4 $
$ T\left(4 V_{1}\right)^{2 / 3}=T^{\prime}\left(32 V_{1}\right)^{2 / 3} $
$\Rightarrow T^{\prime}=\frac{T}{4}$
$W_{\text {adia }}=\frac{2 R\left(T-\frac{T}{4}\right)}{3}=\left(\frac{9 R T}{8}\right)$
$\frac{W_{\text {iso }}}{W_{\text {adia }}}=\frac{8}{9} \ln 4=\frac{16}{9} \ln 2$
$\Rightarrow f=\frac{16}{9}=1.78$