What is ΔU for monoatomic gas adiabatic expansion from V to 27V?

ΔU=−4PV/3. T₂=T₁/9 (from TV^(2/3)=const). ΔU=n(3R/2)(T₁/9-T₁)=−4nRT₁/3=−4PV/3.

What is γ for monoatomic ideal gas?

γ=5/3. Cv=3R/2, Cp=5R/2, γ=5/3.

Monoatomic ideal gas initial P V adiabatic expansion to 27V — change in internal energy

JEERANKERS

PhysicsThermodynamics

QMCQThermodynamics

Initial pressure and volume of a monoatomic ideal gas are $P$ and $V$. Change in internal energy in adiabatic expansion to $V_f=27V$:

A) $-2PV(3\sqrt{3}-1)$ B) $\dfrac{4PV}{3}$ C) $-\dfrac{4PV}{3}$ D) $\dfrac{3PV}{4}$

✅ Correct Answer

C) −4PV/3

✓

Solution

Find T₂ using TV^(γ-1) = constant

Monoatomic: $\gamma=5/3$, so $\gamma-1=2/3$.

$T_1V^{2/3}=T_2(27V)^{2/3}=T_2\cdot9\cdot V^{2/3}$

$T_2=\dfrac{T_1}{9}$

Calculate ΔU

$\Delta U=nC_v(T_2-T_1)=n\cdot\dfrac{3R}{2}\left(\dfrac{T_1}{9}-T_1\right)=-\dfrac{4nRT_1}{3}=-\dfrac{4PV}{3}$

📘 Adiabatic Expansion: TV^(γ-1) = const

For adiabatic process: $TV^{\gamma-1}=\text{const}$. $27^{2/3}=9$. $T_2=T_1/9$. $\Delta U=nC_v\Delta T=-4PV/3$. Negative because gas does work at expense of internal energy.

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Important Concepts

γ for Monoatomicγ=5/3, γ−1=2/3. T·V^(2/3)=const in adiabatic.

27^(2/3)=927=3³, so 27^(1/3)=3, 27^(2/3)=9. Therefore T₂=T₁/9.

ΔU CalculationΔU=n(3R/2)(T₁/9−T₁)=n(3R/2)(−8T₁/9)=−4nRT₁/3=−4PV/3.

Negative ΔUIn adiabatic expansion Q=0: ΔU=−W<0. Gas does positive work, temperature drops.

FAQs

Why TV^(2/3)=const for monoatomic?

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From adiabatic TV^(γ-1)=const with γ=5/3: TV^(5/3-1)=TV^(2/3)=const.

How to compute 27^(2/3)?

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27=3³. 27^(1/3)=3. 27^(2/3)=(27^(1/3))²=3²=9.

Why is ΔU=−4PV/3 and not +4PV/3?

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Gas expands (does positive work). In adiabatic: Q=0, ΔU=−W<0. Internal energy decreases.

Can I use PV^γ=const?

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Yes: P₂=P/27^(5/3)=P/243. W=(P₁V₁-P₂V₂)/(γ-1)=(PV-PV/9)/(2/3)=4PV/3. ΔU=−W=−4PV/3 ✓

Is this from JEE Main 2026?

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Yes, this appeared in JEE Main 2026.

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