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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
$\frac{{{P_0}{T_0}R}}{{{P_0} - \alpha }}$
$\frac{{{P_0}{T_0}R}}{{{P_0} + \alpha }}$
${P_0}{T_0}R\,\ln \,2$
${{P_0}{T_0}R}$
Solution
${P_0}{V_0} = nR{T_0}$
${P_0}V = NRT$
${T_f} = 2{T_0}$
$W = \int {PdV} $
$ = \int {\left( {\frac{{nRT}}{V} + \alpha } \right)dV} $
$PV = nRT + \alpha V$
$\int {PdV = \int\limits_{{T_0}}^{2{T_0}} {nRdT + \int\limits_{{V_1}}^{{V_1}} {\alpha dV} } } $
$ = nR{T_0} + \alpha \,{V_i}$
$ = nR{T_0} + \alpha \left( {\frac{{nR{T_0}}}{{{P_0}}}} \right)$
$ = nR{T_0} \left( {1 + \frac{\alpha }{{{P_0}}}} \right)$
$PV = nRT + \alpha V$
$\int {PdV = \int {nRdT + \int {\alpha dV} } } $
$W = nR{T_0} + \alpha \left[ {\frac{{nR{T_0}}}{{{P_0} – \alpha }}} \right]$
$W = nR{T_0}\left[ {1 + \frac{\alpha }{{{P_0} – \alpha }}} \right]$
$ = n{R_0}{T_0}\left[ {\frac{{{P_0}}}{{{P_0} – \alpha }}} \right]$
$ = \frac{{nR{T_0}{P_0}}}{{{P_0} – \alpha }}$
