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MHT CET 2024 (2 May, Shift 1) · Thermodynamic Processes

Thermodynamic Processes question from MHT CET 2024 (2 May, Shift 1)

Q117MHT CET 2024 · 2 MAY · SHIFT 1Medium

Thermodynamic Processes

One mole of an ideal gas at an initial temperature of ' TT ' K\text{K} does ' 6R6R ' of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is 5/35/3, the final temperature of gas will be (R=8.31 J mole−1 K−1)\left(R = 8.31\text{ J mole}^{-1}\text{ K}^{-1}\right)

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Answer: (d)

  1. Adiabatic work by gas: W=nR(T1−T2)γ−1W = \frac{nR(T_1 - T_2)}{\gamma - 1}, with n=1n = 1, γ−1=23\gamma - 1 = \frac{2}{3}.
  2. 6R=R(T−T2)2/3⇒T−T2=46R = \frac{R(T - T_2)}{2/3} \Rightarrow T - T_2 = 4
  3. Gas does work, so it cools: T2=T−4T_2 = T - 4
  4. So the answer is (d) (T−4)(T - 4) K.
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