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 temperature and pressure of one mole of the gas contained in a cylinder are $T_o$ and $P_o$ respectively. The work done by the gas when its temperature doubles isobarically will be

• A. $\frac{P_o{T}_{o}R}{P_o-\alpha }$
• B. $\frac{P_o{T}_{o}R}{P_o+\alpha }$
• C. $P_o{T}_{o}R\ln 2$
• D. $P_o{T}_{o}R$

Answer 1

The correct option is A $\frac{P_o{T}_{o}R}{P_o-\alpha }$

Isobaric process means pressure will be constant. For one mole gas, $n=1$
If $V_i$ is the initial volume, $P_0{V}_i=RT+\alpha V_i,V_i=\frac{R{T}_0}{P_0-\alpha }$
If $V_f$ is the final volume, $P_0{V}_f=R.2T_0+\alpha V_f,V_f=\frac{2R{T}_0}{P_0-\alpha }$
The work done $={\int }_{V_i}^{V_f}{P}_{0}dV=P_0[V_f-V_i]=\frac{P_{0}R{T}_0}{P_0-\alpha }$