JEE Challenger
More from Thermodynamics
ChemMedAdv 2020 • Paper 1

Work Done Expression for Van der Waals Gas in Thermodynamic Processes

In thermodynamics, the expansion work P−VP-V is defined as w=−∫dVPextw = - \int dV P_{\text{ext}} For a system undergoing a specific transformation, the work done expression is given by w=−∫dV(RTV−b−aV2)w = - \int dV \left( \frac{RT}{V-b} - \frac{a}{V^2} \right) Identify the correct statement(s) regarding the applicability of this formula:

Options

A

A system obeying the van der Waals equation of state.

Correct
B

A process carried out reversibly and isothermally.

Correct
C

A process carried out reversibly and adiabatically.

Correct
D

An irreversible process executed under constant pressure.

Topics & Concepts

ThermodynamicsFirst Law of ThermodynamicsReal Gases

Step-by-Step Solution

To determine the conditions under which the given work done expression is valid, we analyze the definition of P−VP-V work in thermodynamics step-by-step:

  1. General Expression for Work Done: The general formula for expansion or compression work done by or on a thermodynamic system is: w=−∫Pext dVw = -\int P_{\text{ext}} \, dV where PextP_{\text{ext}} is the external opposing pressure and dVdV is the change in volume.

  2. Condition for Reversibility: For a reversible process, the system is in quasi-static equilibrium at every infinitesimal step, meaning the external pressure is virtually equal to the internal pressure of the gas: Pext=PgasP_{\text{ext}} = P_{\text{gas}} Thus, for any reversible process, the work expression becomes: w=−∫Pgas dVw = -\int P_{\text{gas}} \, dV

  3. van der Waals Equation of State: For 1 mole1\text{ mole} of a real gas obeying the van der Waals equation of state: (Pgas+aV2)(V−b)=RT\left(P_{\text{gas}} + \frac{a}{V^2}\right)(V - b) = RT Solving for PgasP_{\text{gas}}: Pgas=RTV−b−aV2P_{\text{gas}} = \frac{RT}{V - b} - \frac{a}{V^2}

Substituting Pext=PgasP_{\text{ext}} = P_{\text{gas}} into the work formula yields: w=−∫(RTV−b−aV2)dVw = -\int \left( \frac{RT}{V - b} - \frac{a}{V^2} \right) dV

  1. Analysis of the Options:
  • Option (A): The formula explicitly uses the expression for PgasP_{\text{gas}} derived from the van der Waals equation of state. Therefore, the system must satisfy the van der Waals equation of state. (Correct)
  • Option (B): For a reversible isothermal process, Pext=PgasP_{\text{ext}} = P_{\text{gas}} holds true, and temperature TT is constant. The given integral expression correctly represents the work done in this process. (Correct)
  • Option (C): For a reversible adiabatic process, Pext=PgasP_{\text{ext}} = P_{\text{gas}} also holds true at every state along the path. Although TT varies as a function of volume VV, the integral expression w=−∫Pgas dV=−∫(RTV−b−aV2)dVw = -\int P_{\text{gas}} \, dV = -\int \left( \frac{RT}{V - b} - \frac{a}{V^2} \right) dV remains definitionally correct and applicable. (Correct)
  • Option (D): In an irreversible process, Pext≠PgasP_{\text{ext}} \neq P_{\text{gas}} during the transition, so PextP_{\text{ext}} cannot be substituted by the state equation Pgas(T,V)P_{\text{gas}}(T,V). Therefore, this equation does not apply to an irreversible process. (Incorrect)

Correct Answers: (A), (B), (C)

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