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…

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 : \(\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\)
  1. 378 K and 500 J
  2. 368 K and 500 J
  3. 348 K and 300 J
  4. 378 K and 300 J

Solution

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

Asked in: JEE Main 2025 (29 Jan Shift 1)

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