Question

If the ratio of specific heat of a gas at constant pressure to that at constant volume is $\gamma$ , the change in internal energy of the mass of gas, when the volume changes from V to 2V at constant pressure P is :

• A. $\frac{R}{(\gamma -1)}$
• B. $PV$
• C. $\frac{PV}{(\gamma -1)}$
• D. $\frac{\gamma PV}{(\gamma -1)}$

Answer 1

The correct option is C $\frac{PV}{(\gamma -1)}$

Using first law we have $\Delta Q=\Delta U+\Delta W$
For an isobaric process we have $\Delta W=PV$
And $\Delta Q$ is given as $n{C}_p\Delta T=n\frac{\gamma R}{\gamma -1}\Delta T$
Using Gas Law equation, we have $PV=nR\Delta T$
Thus, we get $\Delta Q=\frac{\gamma }{\gamma -1}PV$
Hence,
$\Delta U=\frac{\gamma }{\gamma -1}PV-PV=\frac{PV}{\gamma -1}$