Nitrous oxide constant |ThermodynamicsPhysicsJEEMain2022

JEE MAIN 2022

29-06-2022 S2

Question

2.2 g of nitrous oxide $\left(\mathrm{N}_2 \mathrm{O}\right)$ 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, $\Delta \mathrm{U}$ is ' $-x$ ' J . The value of ' $x$ ' is $\_\_\_\_$ . [nearest integer] (Given: atomic mass of $\mathrm{N}=14 \mathrm{~g} \mathrm{~mol}^{-1}$ and of $\mathrm{O}=16 \mathrm{~g} \mathrm{~mol}^{-1}$. Molar heat capacity of $\mathrm{N}_2 \mathrm{O}$ is $\left.100 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)$

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Solution

$\begin{aligned} & \mathrm{N}_2 \mathrm{O} \text { moles }=\frac{2.2}{44}=\frac{1}{20} \\ & \Delta \mathrm{H}=\mathrm{nC}_p \Delta \mathrm{~T}=\frac{1}{20} \times 100(-40)=-200 \mathrm{~J} \\ & \Delta \mathrm{U}=\mathrm{q}_p+\mathrm{w} \\ & \mathrm{W}=-\mathrm{P}_{\text {ext }} \Delta \mathrm{V} \\ & \mathrm{W}=-1 \frac{(167.75-217.1)}{1000} \times 101.3 \mathrm{~J} \\ & \mathrm{~W}=+5 \mathrm{~J} \\ & \Delta \mathrm{U}=-200+5=-195 \mathrm{~J}\end{aligned}$

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JEE Main

Chemistry

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