As we know that, $\Delta S = nC _{ V } \ln \frac{ T _{2}}{ T _{1}}+ nR \ln V _{2} V _{1}$
$\because$ The process is isochoric, i.e., volume is constant.
$\therefore \Delta S = nC _{ V } \ln \frac{ T _{2}}{ T _{1}}$
Given:-
$n =2 moles$
$C _{ P }=\frac{3}{2} R [\because$ the gas is monoatomic $]$
$T _{2}=200^{\circ} C =(200+273) K =473 K$
$T _{1}=300^{\circ} C =(300+273) K =573 K$
$\therefore \Delta S =2 \times \frac{3}{2} R \times \ln \left(\frac{573}{473}\right)$
$\Rightarrow \Delta S =3 R \ln \left(\frac{573}{473}\right)$
Hence the change in entropy of the gas is $3 R \ln \left(\frac{573}{473}\right)$
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An ideal gas undergoes a cyclic transformation starting from the point $\mathrm{A}$ and coming back to the same point by tracing the path $\mathrm{A} \rightarrow \mathrm{B} \rightarrow \mathrm{C} \rightarrow \mathrm{A}$ as shown in the diagram. The total work done in the process is____ $J$.
