Chemistry · Thermodynamics & Thermochemistry

JEE Main 2025 — 29 January, Morning Shift — Question 10

500 J of energy is transferred as heat to 0.5 mol of Argon gas at 298 K and 1.00 atm . The final temperature and the change in internal energy respectively are: Given : $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$

Option A:
348 K and 300 J
Correct
Option B:
378 K and 300 J
Option C:
368 K and 500 J
Option D:
378 K and 500 J

Answer: A

Step-by-step solution

$\quad \mathrm{q}_{\mathrm{p}}=\mathrm{n} \times \mathrm{c}_{\mathrm{p}} \times \Delta \mathrm{T}$ $\Rightarrow 500=0.5 \times \frac{5}{2} \times 8.3\left(\mathrm{~T}_{\mathrm{f}}-298\right)$ $\Rightarrow \mathrm{T}_{\mathrm{f}} \simeq 346.2 \mathrm{~K}$ $\frac{\Delta \mathrm{H}}{\Delta \mathrm{U}}=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\left(\frac{5}{3}\right)$ $\Rightarrow \Delta \mathrm{U}=\frac{3}{5} \times 500=300 \mathrm{~J}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 29 January, Morning Shift
Subject: Chemistry
Chapter: Thermodynamics & Thermochemistry
Topic: Internal Energy and the First Law