For an isothermal reversible expansion of 2 moles of an ideal gas from 10 dm³ to 100 dm³ at 27°C (300 K), what is the entropy change ΔS of the gas (in J K⁻¹)?

Introduction:
This chemistry and thermodynamics question asks you to calculate the entropy change for an isothermal (constant temperature) reversible expansion of an ideal gas. Entropy calculations are common in physical chemistry and chemical thermodynamics, especially for ideal gas processes.

Given Data / Assumptions:

• Number of moles n = 2 mol.
• Initial volume V1 = 10 dm³.
• Final volume V2 = 100 dm³.
• Temperature T = 27°C = 300 K.
• Gas behaves ideally and the process is isothermal and reversible.
• Universal gas constant R ≈ 8.314 J mol⁻¹ K⁻¹.

Concept / Approach:
For an isothermal reversible expansion of an ideal gas, the entropy change of the gas is given by:
ΔS = n * R * ln(V2 / V1)where n is the number of moles, R is the gas constant, and V2/V1 is the ratio of final to initial volume. Because the process is reversible and isothermal, only this simple expression is needed. We then compute ΔS numerically and match it with the closest option.

Step-by-Step Solution:
Step 1: Compute the volume ratio V2 / V1 = 100 / 10 = 10.Step 2: Use the formula ΔS = n * R * ln(V2 / V1).Step 3: Substitute values: ΔS = 2 * 8.314 * ln(10).Step 4: Use ln(10) ≈ 2.3026.Step 5: First multiply 2 * 8.314 = 16.628.Step 6: Now multiply by ln(10): 16.628 * 2.3026 ≈ 38.3 J K⁻¹.Step 7: Therefore, ΔS ≈ 38.3 J K⁻¹ for the gas.

Verification / Alternative check:
We can confirm the magnitude approximately: R is about 8.3, n is 2, so n*R ≈ 16.6. Since ln(10) is about 2.3, the result should be near 16.6 * 2.3 ≈ 38.2. This rough estimate is consistent with the more precise calculation of 38.3 J K⁻¹.

Why Other Options Are Wrong:
42.3 J K⁻¹ and 35.8 J K⁻¹ are somewhat close but do not match the correct product of 16.628 and 2.3026. 32.3 J K⁻¹ and 30.0 J K⁻¹ are noticeably smaller and would correspond to a smaller volume ratio or fewer moles. Only 38.3 J K⁻¹ aligns with the correct calculation from the thermodynamic formula.

Common Pitfalls:
Typical mistakes include using log base 10 instead of natural log, forgetting to convert temperature to Kelvin (although T does not appear explicitly in this formula for ΔS of an ideal gas in an isothermal process), or using V1/V2 instead of V2/V1 in the logarithm. Always use ln(V2/V1) and ensure that volumes are in the same units so their ratio is dimensionless.

Final Answer:
The entropy change for the isothermal reversible expansion is approximately 38.3 J K⁻¹.