JEE Mains · Chemistry · STD 11 - 5. Thermodynamics and thermochemistry
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 : \(\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\)
- A 378 K and 500 J
- B 368 K and 500 J
- C 348 K and 300 J
- D 378 K and 300 J
Answer & Solution
Correct Answer
(C) 348 K and 300 J
Step-by-step Solution
Detailed explanation
\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} \\ &…
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