Chemistry · Chemical Bonding

JEE Advanced 2020 — Paper 1 — Question 19

In thermodynamics, the $\mathrm{P}-\mathrm{V}$ 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

Option A:
system that satisfies the van der Waal's equation of state
Correct
Option B:
process that is reversible and isothermal
Correct
Option C:
process that is reversible and adiabatic
Correct
Option D:
process that is irreversible and at constant pressure

Answer: A, B, C

Step-by-step solution

$W_{\text {rev }}=-\int P d V$

For one mole of a real gas,

$\left(P+\frac{a}{V^{2}}\right)(V-b)=R T$

\begin{gathered} \mathrm{P}=\left(\frac{\mathrm{RT}}{(\mathrm{V}-\mathrm{b})}-\frac{\mathrm{a}}{\mathrm{V}^{2}}\right) …(1) \end{gathered}

So, $W_{\text {rev }}=-\int\left(\frac{R T}{(V-b)}-\frac{a}{V^{2}}\right) d V$

So, above Equation is valid for all reversible processes and for real gases.

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2020
Paper: Paper 1
Subject: Chemistry
Chapter: Chemical Bonding
Topic: Molecular Orbital Theory