In thermodynamics the P-V work done is given by $$\mathrm{w}=-\int \mathrm{dV} \mathrm{P}_{\mathrm{ext}}.$$ For a system undergoing a particular process, the work done is, $$\mathrm{w}=-\int \mathrm{dV}\left(\frac{\mathrm{RT}}{\mathrm{~V}-\mathrm{b}}-\frac{\mathrm{a}}{\mathrm{~V}^{2}}\right).$$ This equation is applicable to a

(A) System that satisfies the van der Waals 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.

JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · States of Matter

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In thermodynamics the P-V work done is given by

w = - \int dV P_{ext} .

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

w = - \int dV (\frac{RT}{V - b} - \frac{a}{V ^{2}}) .

This equation is applicable to a

(A)System that satisfies the van der Waals equation of state.✓ Correct
(B)Process that is reversible and isothermal.✓ Correct
(C)Process that is reversible and adiabatic.✓ Correct
(D)Process that is irreversible and at constant pressure.

Answer: (A), (B), (C)

Solution

The given work expression contains the pressure term from the van der Waals equation:

P = \frac{R T}{V - b} - \frac{a}{V ^{2}}

.

For this expression to be valid:

The system must follow the van der Waals equation of state (option A is correct)
Since $P_{e x t} = P$ , the process must be reversible (eliminates option D)
The expression is valid for any reversible process (isothermal or adiabatic) involving a van der Waals gas

Therefore, options A, B, and C are correct.

JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · One or more correct options

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JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · States of Matter

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