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A gas system is taken through the thermodynamic cyclic
process 1 → 2 → 3 → 1 as shown below. The amount of
heat released by the system is
Options:
process 1 → 2 → 3 → 1 as shown below. The amount of
heat released by the system is
Solution:
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Verified Answer
The correct answer is:
$-\mathrm{P} \frac{\mathrm{V}}{2}$
As process is cyclic
$\begin{aligned}
& \Delta \mathrm{U}=0 \\
& \Delta \mathrm{Q}=\Delta \mathrm{W} \\
& =\text { Area enclosed by curve } \\
& =\frac{1}{2} \times \mathrm{V} \times \mathrm{P}=\frac{\mathrm{PV}}{2}
\end{aligned}$
As process is clockwise, so $\Delta \mathrm{W}$ and $\Delta \mathrm{Q}$ is positive i.e. $\frac{\mathrm{PV}}{2}$ amount of heat is absorbed or $-\frac{\mathrm{PV}}{2}$ amount is released.
$\begin{aligned}
& \Delta \mathrm{U}=0 \\
& \Delta \mathrm{Q}=\Delta \mathrm{W} \\
& =\text { Area enclosed by curve } \\
& =\frac{1}{2} \times \mathrm{V} \times \mathrm{P}=\frac{\mathrm{PV}}{2}
\end{aligned}$
As process is clockwise, so $\Delta \mathrm{W}$ and $\Delta \mathrm{Q}$ is positive i.e. $\frac{\mathrm{PV}}{2}$ amount of heat is absorbed or $-\frac{\mathrm{PV}}{2}$ amount is released.
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