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MHT CET 2026 (19 May, Shift 2) · Thermodynamic Processes

Thermodynamic Processes question from MHT CET 2026 (19 May, Shift 2)

Q27MHT CET 2026 · 19 MAY · SHIFT 2Medium

Thermodynamic Processes

Two identical containers A and B with friction-less pistons contain the same ideal gas at the same temperature and the same volume, VV. The mass of gas in A is mAm_A and that in B is mBm_B. The gas in each cylinder is now allowed to expand isothermally to the same final volume 2V2V. The changes in pressure in A and B are found to be Δp\Delta p and 3Δp3\Delta p respectively. Then

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Answer: (c)

  1. Same VV and TT: initial pressure p∝p \propto number of moles ∝m\propto m.
  2. Isothermal expansion to 2V2V: pf=p2p_f = \frac{p}{2}, so Δp=p2∝m\Delta p = \frac{p}{2} \propto m.
  3. mAmB=Δp3Δp=13⇒mB=3mA\frac{m_A}{m_B} = \frac{\Delta p}{3\Delta p} = \frac{1}{3} \Rightarrow m_B = 3m_A
  4. So the answer is (c) 3mA=mB3m_A = m_B.
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