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

Question

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 $V_{final} = 27 V$ is ______ J. [2026]

Smart Achievers · Accepted Answer

(4)

$P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$

$Here, γ = \frac{5}{3}$

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

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

$Δ U = n C_{v} Δ T = \frac{n R Δ T}{(γ - 1)}$

$Δ U = \frac{P_{2} V_{2} - P_{1} V_{1}}{(γ - 1)}$

$\Rightarrow Δ U = \frac{\frac{P}{3^{5}} \times 3^{3} \cdot V - P V}{(\frac{5}{3} - 1)} = - \frac{4}{3} P V$

Solution

Q.
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 $V_{final} = 27 V$ is ______ J. [2026]

1 $- 2 P V (3 \sqrt{3} - 1)$

2 $\frac{4}{3} P V$

3 $\frac{3}{4} P V$

4 $- \frac{4}{3} P V$

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Solution

Q. 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 Vfinal=27V is ______ J. [2026]

1 -2PV(33-1)

2 43PV

3 34PV

4 -43PV