Verified Solution • 2025–26 Session moderate

QJEEAdv20Ch12d7: JEE Advanced 2020 Chemistry - Class 11 JEEAdv20Ch12d7 Solution: In thermodynamics, the p-V work done is given by For a system…

In thermodynamics, the p-V work done is given by

w=−∫dVpextw = - \int {dV{p_{ext}}}

For a system undergoing a particular process, the work done is

w=−∫dV(RTV−b−aV2)w = - \int {dV\left( {{{RT} \over {V - b}} - {a \over {{V^2}}}} \right)}

This equation is applicable to a

Multiple Choice Options

  • A system that satisfies the van der Walls' equation of state
  • B process that is reversible and isothermal
  • C process that is reversible and adiabatic
  • D process that is irreversible and at constant pressure

Step-by-Step Verified Solution

Given, w=−∫pextdVw = - \int {{p_{ext}}dV}

For 1 mole van der Walls' gas

p=(RTV−b−aV2)p = \left( {{{RT} \over {V - b}} - {a \over {{V^2}}}} \right)

For reversible process, pext = pgas

w=−∫(RTV−b−aV2)dVw = - \int {\left( {{{RT} \over {V - b}} - {a \over {{V^2}}}} \right)dV}

But, it is not applicable for irreversible process which are carried out very fast. So, work done is calculated assuming final pressure remains constant throughout the process. Thus, statement (a), (b) and (c) correct while statement (d) is incorrect.

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