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MHT CET 2026 (16 Apr, Shift 1) · Thermodynamic Processes

Thermodynamic Processes question from MHT CET 2026 (16 Apr, Shift 1)

Q33MHT CET 2026 · 16 APR · SHIFT 1Medium

Thermodynamic Processes

An ideal gas at 27∘27^\circC is compressed adiabatically to (827)\left(\frac{8}{27}\right) of its original volume. If γ=5/3\gamma = 5/3, then the rise in temperature of a gas is

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Answer: (c)

  1. T2=T1(V1V2)γ−1=300(278)2/3T_2 = T_1\left(\frac{V_1}{V_2}\right)^{\gamma - 1} = 300\left(\frac{27}{8}\right)^{2/3}
  2. (278)2/3=(32)2=94\left(\frac{27}{8}\right)^{2/3} = \left(\frac{3}{2}\right)^2 = \frac{9}{4}, so T2=675 KT_2 = 675\text{ K}.
  3. Rise =675−300=375 K= 675 - 300 = 375\text{ K}
  4. So the answer is (c) 375 K.
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