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Q.
Two moles of helium gas undergo a cyclic process as shown in figure. Assuming the gas to be ideal, the net work done by the gas is
Thermodynamics
Solution:
Process $AB$ and $CD -$ isobaric
Workdone $= P \Delta V = nR \Delta T$
$WAB = nR (100)$
$WCD =- nR (100)$
Process $BC$ and $DA \Rightarrow $ isothermal
Work done $= nRT \ln \frac{ P _{1}}{ P _{2}}$
WBC $= nRT \ln \frac{1}{2}=-400 nR \ln \frac{1}{2}$
W $\Delta A = nRT \ln 2=300 nR \ln 2$
Total work done $=100 nR -100 nR +400 R \ln 2-300 nR \ln 2$
$\quad=100 nR \ln 2$
$\quad=200 R \ln 2 \quad \quad( n =2)$
