JEE Main · 2025 · Shift-IeasyTHERMO-148

One mole of an ideal gas expands isothermally and reversibly from 10 dm3 to 20 dm3 at 300 K. U, q and work done in the…

Thermodynamics · Class 11 · JEE Main Previous Year Question

Question

One mole of an ideal gas expands isothermally and reversibly from 10 dm3^3 to 20 dm3^3 at 300 K. ΔU\Delta U, qq and work done in the process respectively are:

Given: R=8.3R = 8.3 JK−1^{-1} and mol−1^{-1}

In 10=2.310 = 2.3, log⁡2=0.30\log 2 = 0.30, log⁡3=0.48\log 3 = 0.48

Options
  1. a

    0,21.840, 21.84 kJ, −1.26-1.26 kJ

  2. b

    0,−17.180, -17.18 kJ, 1.7181.718 J

  3. c

    0,21.840, 21.84 kJ, 21.8421.84 kJ

  4. d

    0,1.7180, 1.718 kJ, −1.718-1.718 kJ

    ✓
Correct Answerd

0,1.7180, 1.718 kJ, −1.718-1.718 kJ

Detailed Solution

🧠 Isothermal expansion of ideal gas: ΔU=0\Delta U = 0, so q=−wq = -w Internal energy of an ideal gas depends only on temperature. Isothermal means TT constant → ΔU=0\Delta U = 0 → heat absorbed equals work done by gas.

🗺️ Calculation w=−nRTln⁡VfVi=−(1)(8.3)(300)(ln⁡2)=−2490×(2.3×0.30)=−2490×0.69w = -nRT\ln\frac{V_f}{V_i} = -(1)(8.3)(300)(\ln 2) = -2490 \times (2.3 \times 0.30) = -2490 \times 0.69 w=−1718 J=−1.718 kJw = -1718\,\text{J} = -1.718\,\text{kJ}

Since ΔU=0\Delta U = 0: q=−w=+1.718 kJq = -w = +1.718\,\text{kJ}

Answer: (d) ΔU=0,  q=+1.718 kJ,  w=−1.718 kJ\boxed{\text{Answer: (d) } \Delta U = 0,\; q = +1.718\,\text{kJ},\; w = -1.718\,\text{kJ}}

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