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JEE Advanced 2022
Paper-1 2022
Question
List I describes thermodynamic processes in four different systems. List II gives the magnitudes (either exactly or as a close approximation) of possible changes in the internal energy of the system due to the process.

Which one of the following options is correct?
Select the correct option:
A
I → T, II → R, III → S, IV → Q
B
I → S, II → P, III → T, IV → P
C
I → P, II → R, III → T, IV → Q
D
I → Q, II → R, III → S, IV → T
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$ \begin{aligned} & \text { (I) } U=M L-P \Delta V \\ & =10^{-3} \times 2250-10^2 \mathrm{kP} \times\left(10^{-3}-10^{-6}\right) \mathrm{m}^3 \\ & =2.25 \mathrm{~kJ}-0.1 \mathrm{~kJ} \\ & =2.15 \mathrm{~kJ} \\ & \mathrm{I} \rightarrow \mathrm{P} \end{aligned} $
(II) $\mathrm{C}_{\mathrm{V}}=\frac{5 \mathrm{R}}{2}$ (rigid diatomic)
For isobasic expansion
$ \begin{aligned} & \mathrm{V} \propto \mathrm{~T} \\ & \frac{\mathrm{~V}_1}{\mathrm{~V}_2}=\frac{\mathrm{T}_1}{\mathrm{~T}_2} \Rightarrow \frac{\mathrm{~V}}{3 \mathrm{~V}}=\frac{500}{\mathrm{~T}_2} \Rightarrow \mathrm{~T}_2=1500 \mathrm{~K} \\ & \Delta \mathrm{U}=\mathrm{nC}_{\mathrm{V}} \Delta \mathrm{~T}=0.2 \times \frac{5 \times 8}{2} \times(1500-500) \mathrm{J} \\ & =4 \mathrm{~kJ} \\ & \mathrm{II} \rightarrow \mathrm{R} \end{aligned} $
(III) Adiabatic expansion $\left(\gamma=\frac{5}{3}\right)$
$ \begin{aligned} & \mathrm{P}_1 . \mathrm{V}_1^\gamma=\mathrm{P}_2 \mathrm{~V}_2^\gamma \Rightarrow(2 \mathrm{kPa}) \times \mathrm{V}_0^{5 / 3}=\mathrm{P}_2 \times\left(\frac{\mathrm{V}_0}{8}\right)^{5 / 3} \\ & \mathrm{P}_2=64 \mathrm{kPa} \\ & \Delta \mathrm{U}=\mathrm{nCV} \Delta \mathrm{~T} \\ & =\frac{3 \mathrm{nR} \Delta \mathrm{~T}}{2}=\frac{3}{2}\left(\mathrm{P}_2 \mathrm{~V}_2-\mathrm{P}_1 \mathrm{~V}_1\right)=\frac{3}{2} \times\left(64 \times \frac{1}{3 \times 8}-2 \times \frac{1}{3}\right) \\ & =\frac{3}{2} \times\left(\frac{8}{3}-\frac{2}{3}\right)=3 \mathrm{~kJ} \\ & \mathrm{III} \rightarrow \mathrm{~T} \end{aligned} $
(IV) For isobaric expansion,
$ \begin{aligned} & \Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{~T}=\frac{7}{2} \mathrm{nR} \Delta \mathrm{~T} \\ & \Delta \mathrm{Q}=\mathrm{nC}_{\mathrm{P}} \Delta \mathrm{~T}=\frac{9}{2} \mathrm{nR} \Delta \mathrm{~T} \\ & \frac{\Delta \mathrm{U}}{\Delta \mathrm{Q}}=\frac{7}{9} \\ & \Delta \mathrm{U}=\frac{7}{9} \Delta \mathrm{Q}=7 \mathrm{~kJ} \\ & \mathrm{IV} \rightarrow \mathrm{Q} \end{aligned} $
Question Tags
JEE Advance
Physics
Easy
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