JEE Main · 2025 · Shift-IhardTHERMO-143

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…

Thermodynamics · Class 11 · JEE Main Previous Year Question

Question

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=8.3R = 8.3 J K−1^{-1} mol−1^{-1})

Options
  1. a

    348 K and 300 J

    ✓
  2. b

    378 K and 300 J

  3. c

    368 K and 500 J

  4. d

    378 K and 500 J

Correct Answera

348 K and 300 J

Detailed Solution

🧠 Monoatomic ideal gas at constant pressure: CP=52RC_P = \frac{5}{2}R and CV=32RC_V = \frac{3}{2}R Heat at constant pressure heats the gas AND does PΔVP\Delta V work. Only 35\frac{3}{5} of the heat input raises internal energy.

🗺️ Calculation n=0.5 moln = 0.5\,\text{mol}, CP=52R=52×8.3=20.75 J K−1mol−1C_P = \frac{5}{2}R = \frac{5}{2} \times 8.3 = 20.75\,\text{J K}^{-1}\text{mol}^{-1}

Temperature rise: qP=nCPΔT⇒500=0.5×20.75×ΔTq_P = nC_P\Delta T \Rightarrow 500 = 0.5 \times 20.75 \times \Delta T ΔT=50010.375=48.2 K\Delta T = \frac{500}{10.375} = 48.2\,\text{K}

Final temperature: Tf=298+48.2=346.2 K≈348 KT_f = 298 + 48.2 = 346.2\,\text{K} \approx 348\,\text{K}

ΔU=nCVΔT=0.5×32(8.3)×48.2=0.5×12.45×48.2=300 J\Delta U = nC_V\Delta T = 0.5 \times \frac{3}{2}(8.3) \times 48.2 = 0.5 \times 12.45 \times 48.2 = 300\,\text{J}

Answer: (a) 348 K and 300 J\boxed{\text{Answer: (a) 348 K and 300 J}}

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