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The change in internal energy of the mass of gas, when the volume changes from V to 2V at constant pressure P is (\(γ\) is the ratio of specific heat at constant pressure to that of volume) (\(γ = \frac{C_p}{C_v}\))
The change in internal energy of the mass of gas, when the volume changes from V to 2V at constant pressure P is (\(γ\) is the ratio of specific heat at constant pressure to that of volume) (\(γ = \frac{C_p}{C_v}\))
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Use the ideal gas law for the temperature change, then dU = n Cv dT with Cv = R/(gamma - 1).
Updated On: Oct 1, 2026
- \(\frac{PV}{γ-1}\)
- \(\frac{PV}{γ+1}\)
- \(\frac{(γ-1)}{PV}\)
- \(\frac{(γ+1)}{PV}\)
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The Correct Option is A
Solution and Explanation
Step 1: Use the first law.
$Q = \Delta U + W$. At constant pressure, $Q = nC_p\Delta T$ and $W = P\Delta V = PV$.
Step 2: Use the ratio.
$\Delta U = Q/\gamma$ because $C_v/C_p = 1/\gamma$. Also $W = Q - \Delta U = \Delta U(\gamma - 1)$.
Step 3: Solve.
So $PV = \Delta U(\gamma - 1)$, giving $\Delta U = \dfrac{PV}{\gamma - 1}$.
Final Answer:
Option (A). \[ \boxed{\frac{PV}{\gamma - 1}} \]
$Q = \Delta U + W$. At constant pressure, $Q = nC_p\Delta T$ and $W = P\Delta V = PV$.
Step 2: Use the ratio.
$\Delta U = Q/\gamma$ because $C_v/C_p = 1/\gamma$. Also $W = Q - \Delta U = \Delta U(\gamma - 1)$.
Step 3: Solve.
So $PV = \Delta U(\gamma - 1)$, giving $\Delta U = \dfrac{PV}{\gamma - 1}$.
Final Answer:
Option (A). \[ \boxed{\frac{PV}{\gamma - 1}} \]
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