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JEE Mains · Physics · STD 11 - Thermodynamics
  • JEE Mains 2026
  • Medium
  • MCQ

Initial pressure and volume of a monoatomic ideal gas are \(P\) and \(V\). The change in internal energy of this gas in adiabatic expansion to volume \(V_{final}=27V\) is ________ J.

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Answer & Solution

Correct Answer

(C) \(-\dfrac{4}{3}PV\)

Step-by-step Solution

Detailed explanation

For a monoatomic ideal gas, the ratio of specific heats is \(\gamma = \dfrac{5}{3}\).

In an adiabatic process, \(PV^{\gamma} = \text{constant}\).

\(P_1 V_1^{\gamma} = P_2 V_2^{\gamma}\)

\(P V^{5/3} = P_2 (27V)^{5/3}\)

\(P_2 = P \left(\dfrac{1}{27}\right)^{5/3} = \dfrac{P}{243}\)

The change in internal energy is given by:

\(\Delta U = \dfrac{P_2 V_2 - P_1 V_1}{\gamma - 1}\)

Substituting the values:

\(\Delta U = \dfrac{\left(\dfrac{P}{243}\right)(27V) - PV}{\dfrac{5}{3} - 1}\)

\(\Delta U = \dfrac{\dfrac{PV}{9} - PV}{\dfrac{2}{3}}\)

\(\Delta U = \dfrac{-\dfrac{8}{9}PV}{\dfrac{2}{3}} = -\dfrac{4}{3}PV\)

Answer: \(-\dfrac{4}{3}PV\)
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