Question

The equation of state for a gas is given by $PV=nRT+\alpha V$, where $n$ is the number of moles and $\alpha$ is a positive constant. The initial pressure and temperature of 1 mole of the gas contained in a cylinder are $P_0$ and $T_0$ respectively. The work done by the gas when its temperature doubles isobarically is

• A. $W=\frac{P_{0}R{T}_0}{P_0-\alpha }$
• B. $W=\frac{R{T}_0}{P_0-\alpha }$
• C. $W=\frac{2P_{0}R{T}_0}{P_0-\alpha }$
• D. $W=\frac{P_{0}R{T}_0}{2(P-\alpha )}$

Answer 1

The correct option is A $W=\frac{P_{0}R{T}_0}{P_0-\alpha }$

Work done for isobaric process,
$\Delta W=P\Delta V......\left(\text{I}\right)$
From question,
$PV=nRT+\alpha V$
$\Rightarrow P\Delta V=nR\Delta T+\alpha \Delta V$
[Here, $P,n,R\text{and}\alpha$ are constants]
$\Rightarrow P\Delta V-\alpha \Delta V=nR\Delta T$
$\Rightarrow \Delta V=\frac{nR\Delta T}{P-\alpha }$
Putting the value of $\Delta V$ in Eq.(I),
$\Rightarrow \Delta W=P\Delta V=\frac{PnR\Delta T}{P-\alpha }$
[Given $n=1\text{mole},P_i=P_0,T_i=T_0$]
$\Delta T=2T_0-T_0=T_0$
$\Rightarrow \Delta W=P\Delta V=\frac{P_{0}R{T}_0}{P_0-\alpha }$
Hence, option (a) is the correct answer.