JEE Main 2025 Jan 29 Shift 1, Chemistry Q13: Calorimetry

JEE Main2025Jan 29, Shift 1Chemistry

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 : $R = 8.3 J K^{- 1} mo l^{- 1}$

A
346 K and 300 J
B
378 K and 300 J
C
368 K and 500 J
D
378 K and 500 J

Solution

Argon is monoatomic, so $C_{p, m} = \frac{5}{2} R$ and $C_{v, m} = \frac{3}{2} R$ .

At constant pressure $q_{p} = n C_{p, m} Δ T$ , giving

$Δ T = \frac{500}{0.5 \times ( 5/2 ) ( 8.3 )} = \frac{500}{10.375} \approx 48.2 K \Rightarrow T_{f} \approx 346 K .$

For $Δ U$ at constant pressure: $Δ U = n C_{v, m} Δ T = \frac{C _{v, m}}{C _{p, m}} q_{p} = \frac{3}{5} \times 500 = 300 J .$

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JEE Main 2025 Jan 29 Shift 1, Chemistry Q13: Calorimetry

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