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JEE Mains · Physics · STD 11 - 11. thermodynamics

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

  1. A \(\frac{{{P_0}{T_0}R}}{{{P_0} - \alpha }}\)
  2. B \(\frac{{{P_0}{T_0}R}}{{{P_0} + \alpha }}\)
  3. C \({P_0}{T_0}R\,\ln \,2\)
  4. D \({{P_0}{T_0}R}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{{{P_0}{T_0}R}}{{{P_0} - \alpha }}\)

Step-by-step Solution

Detailed explanation

\({P_0}{V_0} = nR{T_0}\) \({P_0}V = NRT\) \({T_f} = 2{T_0}\) \(W = \int {PdV} \) \( = \int {\left( {\frac{{nRT}}{V} + \alpha } \right)dV} \) \(PV = nRT + \alpha V\) \(\int {PdV = \int\limits_{{T_0}}^{2{T_0}} {nRdT + \int\limits_{{V_1}}^{{V_1}} {\alpha dV} } } \)…
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