JEE Main 2022 — Chemistry Physical Chemistry

Molecular weight of nitrous oxide $\left(\mathrm{N}_2 \mathrm{O}\right)=44 \mathrm{~g} / \mathrm{mol}$(N2​O)=44 g/mol Number of moles of $\mathrm{N}_2 \mathrm{O}=\frac{2.2}{44} \quad 0.05 \mathrm{~mol}$N2​O=442.2​0.05 mol Specific heat capacity of $\mathrm{N}_2 \mathrm{O}=880 \mathrm{~J} / \mathrm{g} . \mathrm{K}$N2​O=880 J/g.K $\therefore$∴ Heat release on cooling from $310 \mathrm{~K}$310 K to $270 \mathrm{~K}$270 K $q=-2.2 \times 880 \times 40 \mathrm{~K}$q=−2.2×880×40 K $q=-77.44 \mathrm{KJ}$q=−77.44KJ Negative sign indicates heat released. $\therefore$∴ Work done by surrounding on system to compares the gas from $217.1 \mathrm{ml}$217.1ml to $167.75 \mathrm{ml}$167.75ml at constant pressure. It means work done by surrounding is irreversible. $\begin{aligned} & \therefore W= P \times \Delta V \\ & =1.01 \times 10^5 \times(167.75 \quad 217.1 \mathrm{ml} l) \\ & =1.01 \times 10^5 \times 0.049 J \\ & W=4.95 k . J \end{aligned}$​∴W=P×ΔV=1.01×105×(167.75217.1mll)=1.01×105×0.049JW=4.95k.J​ Using 1st Law of thermodynamics $\begin{aligned} & \Delta U=q+W \\ & \Delta U=-77.44 K J + 4.95 K J \\ & \Delta U=-72.49 K J \end{aligned}$​ΔU=q+WΔU=−77.44KJ+4.95KJΔU=−72.49KJ​.

$$\therefore x = 72.$$∴x=72.