Chemistry · States of Matter - Gaseous State

JEE Advanced 2023 — Paper 1 — Question 39

A gas has a compressibility factor of 0.5 and a molar volume of $0.4 \mathrm{dma}^{\text {nol }}{ }^{-1}$ at a temperature of 800 K and pressure $\mathbf{x}$ atm. If it shows ideal gas behaviour at the same temperature and pressure, the molar volume will be $\mathbf{y}$ dna nol ${ }^{-1}$ . The value of $\mathbf{x} / \mathbf{y}$ is $\_\_\_\_$ - [Use: Gas constant, $\mathrm{R}=8 \times 10^{-2} \mathrm{~L}$ atm $\mathrm{K}^{-1}$ nol ${ }^{-1}$ ]

Answer: 100

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{Z}=0.5$ $\mathrm{V}=0.4 \mathrm{dm}^{3} \mathrm{~mol}^{-1}$ $\mathrm{T}=800 \mathrm{~K}$ $\mathrm{P}=\mathrm{x}$ $\mathrm{Z}=\frac{\mathrm{PV}}{\mathrm{RT}}=0.5=\frac{\mathrm{x} \times 0.4}{0.08 \times 800}$ $\mathrm{X}=80$ When $\mathrm{Z}=1$ , Ideal condition molar volume $\mathrm{y} \mathrm{dm}^{3}$ $Z=\frac{P V}{R T}$ $1=\frac{80 \times y}{0.08 \times 800}$ $\mathrm{y}=0.8$ $\frac{\mathrm{x}}{\mathrm{y}}=\frac{80}{0.8}=100$

Answer key and solution verified before publishing.

Practise States of Matter - Gaseous State

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Advanced 2023
Paper: Paper 1
Subject: Chemistry
Chapter: States of Matter - Gaseous State
Topic: Real Gases + Compressibility Factor