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Chemistry Question 59 – NEET-UG 2026

Two moles of an ideal gas undergo free expansion from 10 L to 100 L at 300 K. The values of ΔSsystem and ΔSsurroundings are (R is universal gas constant)

Entropy is a measure of the disorder or randomness of a system. For an ideal gas, entropy change depends on changes in volume and temperature.

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Ninja StrategyFree Expansion Properties

Recognize that free expansion of an ideal gas is an irreversible process where ΔSsystem>0 and qsurroundings=0, leading to ΔSsurroundings=0.

Step 1: Calculate ΔSsystem for isothermal expansion✦ Active

For an ideal gas undergoing isothermal expansion, the entropy change of the system is given by the formula ΔSsystem=nRln(V2V1). Given n=2 moles, V1=10 L, and V2=100 L.

ΔSsystem=2Rln(100 L10 L)=2Rln(10) ΔSsystem=2R×2.303=4.606R
Step 2: Calculate ΔSsurroundings for free expansion○ Expand

Free expansion implies that the external pressure is zero (Pext=0). For an ideal gas, this means no work is done (W=0) and no heat is exchanged with the surroundings (q=0). Therefore, the heat absorbed by the surroundings (qsurroundings) is zero.

ΔSsurroundings=qsurroundingsTsurroundings=0300 K=0
💡 Teacher's Secret Hint

Remember that for a free expansion of an ideal gas, q=0 for the system, and thus qsurroundings=0.

Step 3: Combine the results○ Expand

Based on the calculations, ΔSsystem=4.606R and ΔSsurroundings=0. This corresponds to option (4).

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