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 a mass of gas, when the volume changes from $V$ to $3V$ under constant pressure $P$, is

• A. $\frac{R}{(\gamma -1)}$
• B. $PV$
• C. $\frac{2PV}{\gamma -1}$
• D. $\frac{\gamma PV}{\gamma -1}$

Answer 1

The correct option is C $\frac{2PV}{\gamma -1}$

We know, at constant pressure change in internal energy is given as:
$△U=nCv△T$

Now $\frac{Cp}{Cv}=\gamma$ and $C_p-C_v=R$
so, $1+\frac{R}{Cv}=\gamma$
$Cv=\left(\frac{R}{\gamma -1}\right)$

putting value of $C_v$ :
$△U=nCv△T=n\left(\frac{R}{\gamma -1}\right)△TusingP△V=nR△T\text{under constant P,}\Rightarrow △U=\frac{P△V}{(\gamma -1)}=\frac{P(3V-V)}{\gamma -1}=\frac{2PV}{(\gamma -1)}$