The equation of state for a gas is given by PV = nRT + V, where n… - Vidyadip

MCQ

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

✓
$\frac{{{P_0}{T_0}R}}{{{P_0} - \alpha }}$
B
$\frac{{{P_0}{T_0}R}}{{{P_0} + \alpha }}$
C
${P_0}{T_0}R\,\ln \,2$
D
${{P_0}{T_0}R}$

✓

Answer

Correct option: A.

$\frac{{{P_0}{T_0}R}}{{{P_0} - \alpha }}$

a
${P_0}{V_0} = nR{T_0}$

${P_0}V = NRT$

${T_f} = 2{T_0}$

$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 }}$

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