Question:medium
The change in the internal energy of the mass of gas, when the volume changes from V to 2 V at constant pressure P is \(\left(\gamma=\frac{\mathrm{Cp}}{\mathrm{Cv}}\right)\)
The change in the internal energy of the mass of gas, when the volume changes from V to 2 V at constant pressure P is \(\left(\gamma=\frac{\mathrm{Cp}}{\mathrm{Cv}}\right)\)
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Use the internal energy change for an ideal gas and the work done at constant pressure to express it in terms of P, V and gamma.
Updated On: Nov 13, 2026
- \(\frac{\mathrm{V}}{\mathrm{P}(\gamma -1)}\)
- \(\frac{\mathrm{P}}{\mathrm{V}(\gamma -1)}\)
- \(\frac{\mathrm{PV}}{\gamma +1}\)
- \(\frac{\mathrm{PV}}{\gamma -1}\)
Show Solution
The Correct Option is D
Solution and Explanation
Step 1: First law
$Q=\Delta U+W$ with $W=P\Delta V=PV$.
Step 2: Heat at constant pressure
$Q=nC_P\Delta T=\gamma\,\Delta U$.
Step 3: Solve
$\gamma\Delta U=\Delta U+PV\Rightarrow\Delta U=\dfrac{PV}{\gamma-1}$.
Final Answer:
The internal energy changes by $\dfrac{PV}{\gamma-1}$. \[ \boxed{\text{(D) }\dfrac{\mathrm{PV}}{\gamma -1}} \]
$Q=\Delta U+W$ with $W=P\Delta V=PV$.
Step 2: Heat at constant pressure
$Q=nC_P\Delta T=\gamma\,\Delta U$.
Step 3: Solve
$\gamma\Delta U=\Delta U+PV\Rightarrow\Delta U=\dfrac{PV}{\gamma-1}$.
Final Answer:
The internal energy changes by $\dfrac{PV}{\gamma-1}$. \[ \boxed{\text{(D) }\dfrac{\mathrm{PV}}{\gamma -1}} \]
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