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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 $W_{\text {adia, }}$ respectively. If the ratio $\frac{W_{\text {iso }}}{W_{\text {adia }}}=f \ln 2$, then $f$ is. . . . . . . .
$1.78$
$1.80$
$1.85$
$1.90$
Solution
$\frac{ P _1}{4}\left(4 V _1\right)^{5 / 3}= P _2\left(32 v _1\right)^{5 / 3}$
$P _2=\frac{ P _1}{4}\left(\frac{1}{8}\right)^{5 / 3}=\frac{ P _1}{128}$
$W _{ adi }=\frac{ P _1 V _1- P _2 V _2}{\gamma-1}=\frac{ P _1 V _1-\frac{P_1}{128}\left(32 V _1\right)}{\frac{5}{3}-1}$
$=\frac{P_1 V_1(3 / 4)}{2 / 3}=\frac{9}{8} P_1 V_1$
$W _{ iso } \quad= P _1 V _1 \operatorname{In}\left(\frac{4 V _1}{ V _1}\right)=2 P _1 V _1 \ln 2$
$\frac{ W _{ iso }}{ W _{\text {zdio }}}=\frac{2 P _1 V _1 \ln 2}{\frac{9}{8} P _1 V _1}=\frac{16}{9} \ln 2= f \ln 2$
$f =\frac{16}{9}=1.7778 \approx 1.78$
