Physics Q36 – JEE-MAIN 2026

Home Physics ThermodynamicsQuestion #516

Initial pressure and volume of a monatomic ideal gas are

P

and

V

. The change in internal energy of this gas in adiabatic expansion to volume

V_{f i n a l} = 27 V

is _______ J.

Clue 1 of 3: Adiabatic Process Properties

For an adiabatic process, there is no heat exchange with the surroundings, meaning $Q = 0$ . The change in internal energy is solely due to the work done.

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Step 1: Determine the final pressure (

P_{2}

) using the adiabatic process equation.✦ Active

For an adiabatic process, $P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$ . Given initial pressure $P_{1} = P$ , initial volume $V_{1} = V$ , and final volume $V_{2} = 27 V$ . For a monatomic ideal gas, the adiabatic index $γ = \frac{5}{3}$ .

P V^{5 / 3} = P_{2} (27 V)^{5 / 3}

\Rightarrow P_{2} = P {(\frac{V}{27 V})}^{5 / 3} = P {(\frac{1}{27})}^{5 / 3}

\Rightarrow P_{2} = P {({(\frac{1}{3})}^{3})}^{5 / 3} = P {(\frac{1}{3})}^{5} = \frac{P}{243}

Thus, the final pressure is $P_{2} = \frac{P}{243}$ .

Step 2: Calculate the change in internal energy (

Δ U

).○ Expand

The change in internal energy for an ideal gas is given by $Δ U = \frac{P_{2} V_{2} - P_{1} V_{1}}{γ - 1}$ . Substitute the initial and final state variables and $γ$ :

Δ U = \frac{(\frac{P}{243}) (27 V) - P V}{\frac{5}{3} - 1}

\Rightarrow Δ U = \frac{\frac{27 P V}{243} - P V}{\frac{2}{3}} = \frac{\frac{P V}{9} - P V}{\frac{2}{3}}

\Rightarrow Δ U = \frac{P V (\frac{1}{9} - 1)}{\frac{2}{3}} = \frac{P V (- \frac{8}{9})}{\frac{2}{3}}

\Rightarrow Δ U = P V (- \frac{8}{9}) (\frac{3}{2}) = P V (- \frac{4}{3})

The change in internal energy is $Δ U = - \frac{4}{3} P V$ .

💡 Teacher's Secret Hint

Remember that for an expansion, work is done by the gas, so the internal energy decreases, resulting in a negative change in internal energy.

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Physics Question 36 – JEE-MAIN 2026

Clue 1 of 3: Adiabatic Process Properties

🧠 Full Solution Path