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Question:

The entropy change involved in the isothermal reversible expansion of 2 mole of an ideal gas from a volume of 10dm³ to a volume of 100dm³ at 27°C is:

35.8 Jmol⁻¹K⁻¹

42.3 Jmol⁻¹K⁻¹

38.3 Jmol⁻¹K⁻¹

32.3 Jmol⁻¹K⁻¹

Solution:

ΔS = Qrev/T
For isothermal reversible expansion of an ideal gas, Qrev = nRTln(V2/V1)
Therefore, ΔS = nRln(V2/V1)
Given:
n = 2 moles
R = 8.314 Jmol⁻¹K⁻¹
V1 = 10 dm³
V2 = 100 dm³
T = 27°C = 300 K
ΔS = 2 × 8.314 × ln(100/10)
ΔS = 16.628 × ln(10)
ΔS = 16.628 × 2.303
ΔS = 38.29 Jmol⁻¹K⁻¹ ≈ 38.3 Jmol⁻¹K⁻¹
Therefore, the entropy change involved in the isothermal reversible expansion of 2 mole of an ideal gas from a volume of 10dm³ to a volume of 100dm³ at 27°C is approximately 38.3 Jmol⁻¹K⁻¹.