JEE Main 2025 29 January Morning Shift
Exam JEE Main
Year 2025
Date 29 January
Shift Morning Shift
Subject Chemistry
Chapter Thermodynamics (C) JEE Main 2025 · 29 January · Morning Shift · Chemistry · Thermodynamics (C)
500 J of energy is transferred as heat to 0.5 mol of Argon gas at...
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})
Options A 378 K and 500 J B 368 K and 500 J C 348 K and 300 J D 378 K and 300 J Correct answer C. 348 K and 300 J
Step-by-step solution
q p = n × c p × Δ T ⇒ 500 = 0.5 × 5 2 × 8.3 ( T f − 298 ) ⇒ T f ≃ 346.2 K Δ H Δ U = C p C v = ( 5 3 ) ⇒ Δ U = 3 5 × 500 = 300 J \begin{aligned} & \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}\end{aligned} q p = n × c p × Δ T ⇒ 500 = 0.5 × 2 5 × 8.3 ( T f − 298 ) ⇒ T f ≃ 346.2 K Δ U Δ H = C v C p = ( 3 5 ) ⇒ Δ U = 5 3 × 500 = 300 J
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