- JEE Mains 2014
- Hard
- 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
Answer & Solution
(A) \(\frac{{{P_0}{T_0}R}}{{{P_0} - \alpha }}\)
Step-by-step Solution
Detailed explanation
\({P_0}V = NRT\)
\({T_f} = 2{T_0}\)
\(W = \int {PdV} \)
\( = \int {\left( {\frac{{nRT}}{V} + \alpha } \right)dV} \)
\(PV = nRT + \alpha V\)
\(\int {PdV = \int\limits_{{T_0}}^{2{T_0}} {nRdT + \int\limits_{{V_1}}^{{V_1}} {\alpha dV} } } \)
\( = nR{T_0} + \alpha \,{V_i}\)
\( = nR{T_0} + \alpha \left( {\frac{{nR{T_0}}}{{{P_0}}}} \right)\)
\( = nR{T_0} \left( {1 + \frac{\alpha }{{{P_0}}}} \right)\)
\(PV = nRT + \alpha V\)
\(\int {PdV = \int {nRdT + \int {\alpha dV} } } \)
\(W = nR{T_0} + \alpha \left[ {\frac{{nR{T_0}}}{{{P_0} - \alpha }}} \right]\)
\(W = nR{T_0}\left[ {1 + \frac{\alpha }{{{P_0} - \alpha }}} \right]\)
\( = n{R_0}{T_0}\left[ {\frac{{{P_0}}}{{{P_0} - \alpha }}} \right]\)
\( = \frac{{nR{T_0}{P_0}}}{{{P_0} - \alpha }}\)
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