Consider one mole of helium gas enclosed in a | vedclass

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Question

$\frac{ P _1}{4}\left(4 V _1\right)^{5 / 3}= P _2\left(32 v _1\right)^{5 / 3}$

$P _2=\frac{ P _1}{4}\left(\frac{1}{8}\right)^{5 / 3}=\frac{ P _1}{1

Vedclass · Accepted Answer

$1.78$

Vedclass · Answer

$\frac{ P _1}{4}\left(4 V _1ight)^{5 / 3}= P _2\left(32 v _1ight)^{5 / 3}$

$P _2=\frac{ P _1}{4}\left(\frac{1}{8}ight)^{5 / 3}=\frac{ P _1}{128}$

$W _{ adi }=\frac{ P _1 V _1- P _2 V _2}{\gamma-1}=\frac{ P _1 V _1-\frac{P_1}{128}\left(32 V _1ight)}{\frac{5}{3}-1}$

$=\frac{P_1 V_1(3 / 4)}{2 / 3}=\frac{9}{8} P_1 V_1$

$W _{ iso } \quad= P _1 V _1 \operatorname{In}\left(\frac{4 V _1}{ V _1}ight)=2 P _1 V _1 \ln 2$

$\frac{ W _{ iso }}{ W _{	ext {zdio }}}=\frac{2 P _1 V _1 \ln 2}{\frac{9}{8} P _1 V _1}=\frac{16}{9} \ln 2= f \ln 2$

$f =\frac{16}{9}=1.7778 \approx 1.78$

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