Q22 / 90 · JEE Main 2022 (29 Jun Shift 2) · Chemistry

2.2 g of nitrous oxide N2O gas is cooled at a constant pressure of 1 atm from 310 K to 270 K causing the compression of the gas from 217.1 mL to 167.75 mL. The change in internal energy of the process, △U is '-x'J. The value of 'x' is _____.

[nearest integer]

(Given: atomic mass of N=14 g mol-1 and of O=16 g mol-1. Molar heat capacity of N2O is 100 JK-1 mol-1)

Correct answer: 72

Solution

Molecular weight of nitrous oxide \(\left(\mathrm{N}_2 \mathrm{O}\right)=44 \mathrm{~g} / \mathrm{mol}\) Number of moles of \(\mathrm{N}_2 \mathrm{O}=\frac{2.2}{44} \quad 0.05 \mathrm{~mol}\) Specific heat capacity of \(\mathrm{N}_2 \mathrm{O}=880 \mathrm{~J} / \mathrm{g} . \mathrm{K}\) \(\therefore\) Heat release on cooling from \(310 \mathrm{~K}\) to \(270 \mathrm{~K}\) \(q=-2.2 \times 880 \times 40 \mathrm{~K}\) \(q=-77.44 \mathrm{KJ}\) Negative sign indicates heat released. \(\therefore\) Work done by surrounding on system to compares the gas from \(217.1 \mathrm{ml}\) to \(167.75 \mathrm{ml}\) at constant pressure. It means work done by surrounding is irreversible. \(\begin{aligned} & \therefore W= P \times \Delta V \\ & =1.01 \times 10^5 \times(167.75 \quad 217.1 \mathrm{ml} l) \\ & =1.01 \times 10^5 \times 0.049 J \\ & W=4.95 k . J \end{aligned}\) Using 1st Law of thermodynamics \(\begin{aligned} & \Delta U=q+W \\ & \Delta U=-77.44 K J + 4.95 K J \\ & \Delta U=-72.49 K J \end{aligned}\).
$\therefore x = 72.$

Hence, the correct answer is 72.

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