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JEE Main 2026 (8 Apr, Shift 2) · Thermodynamic Processes

Thermodynamic Processes question from JEE Main 2026 (8 Apr, Shift 2)

Q38JEE Main 2026 · 8 APR · SHIFT 2Medium

Thermodynamic Processes

Initial pressure and volume of a monoatomic ideal gas are PP and VV. The change in internal energy of this gas in adiabatic expansion to volume VVfinal =27 V= 27\,V is ______ J.

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Answer: (c)

  1. Monoatomic: γ=53\gamma = \frac{5}{3}; adiabatic T2=T1(127)2/3=T19T_2 = T_1\left(\frac{1}{27}\right)^{2/3} = \frac{T_1}{9}.
  2. ΔU=32nR(T2−T1)\Delta U = \frac{3}{2}nR(T_2 - T_1) and nRT1=PVnRT_1 = PV.
  3. ΔU=32PV(19−1)=−43PV\Delta U = \frac{3}{2}PV\left(\frac{1}{9} - 1\right) = -\frac{4}{3}PV
  4. So the answer is (c) −43PV-\frac{4}{3}PV.
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