JEE Main · 2019 · Shift-IImediumTHERMO-051

An ideal gas undergoes isothermal compression from 5\,m3 to 1\,m3 against a constant external pressure of 4\,N m-2.…

Thermodynamics · Class 11 · JEE Main Previous Year Question

Question

An ideal gas undergoes isothermal compression from 5 m35\,\text{m}^3 to 1 m31\,\text{m}^3 against a constant external pressure of 4 N m−24\,\text{N m}^{-2}. Heat released = 24 J mol−1K−124\,\text{J mol}^{-1}\text{K}^{-1} & is used to increase the pressure of 1 mol Al. Temperature rise of Al increases by:

Options
  1. a

    23 K\frac{2}{3}\,\text{K}

    ✓
  2. b

    1 K1\,\text{K}

  3. c

    32 K\frac{3}{2}\,\text{K}

  4. d

    2 K2\,\text{K}

Correct Answera

23 K\frac{2}{3}\,\text{K}

Detailed Solution

🧠 Isothermal compression: ΔU=0\Delta U = 0, heat released = work done on gas Compression does work on the gas; at constant temperature, that energy leaves as heat. That heat then warms the aluminium.

🗺️ Calculation Work done on gas (constant external pressure): won=−Pext(Vf−Vi)=−4 N m−2×(1−5) m3=+16 Jw_{\text{on}} = -P_{\text{ext}}(V_f - V_i) = -4\text{ N m}^{-2} \times (1 - 5)\text{ m}^3 = +16\text{ J}

Heat released by gas = 16 J (flows into Al).

Temperature rise of 1 mol Al (Cv=24 J mol−1K−1C_v = 24\text{ J mol}^{-1}\text{K}^{-1}): ΔT=QnCv=161×24=23 K\Delta T = \frac{Q}{nC_v} = \frac{16}{1 \times 24} = \frac{2}{3}\text{ K}

⚠️ Trap: This is compression (volume decreases from 5 m³ to 1 m³), so ΔV<0\Delta V < 0 and won>0w_{\text{on}} > 0. The heat released is only 16 J — a very small external pressure and large volume change still gives modest energy because 4 N/m² is tiny.

Answer: (a)\boxed{\text{Answer: (a)}}

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