Final answer:
The entropy change for the isothermal expansion of two moles of an ideal gas from 10 dm³ to 20 dm³ at 300 K is calculated using the formula ΔS = nRln(V2/V1). Upon substituting the values, the entropy change is not exactly one of the provided options. Therefore, the closest answer provided is 20 J/K, which is for learning purposes and not the precise value.
Step-by-step explanation:
The student has asked to calculate the entropy change for two moles of an ideal gas that expands isothermally and reversibly from a volume of 10 dm3 to 20 dm3 at 300 K. The formula to calculate the entropy change (ΔS) for an isothermal process for an ideal gas is ΔS = nRln(V2/V1), where n is the number of moles, R is the ideal gas constant (8.314 J/(mol·K)), V1 is the initial volume, and V2 is the final volume. Substituting the values, we get ΔS = 2 mol * 8.314 J/(mol·K) * ln(20 dm3/10 dm3) = 2 mol * 8.314 J/(mol·K) * ln(2) ≈ 11.53 J/K. Since this value is not listed in the options provided, there might be a mistake in the question or the options provided. The correct answer to an entropy change in an ideal gas during isothermal expansion would typically be a positive value, indicating an increase in entropy. Therefore, the closest correct option from the ones given, assuming a calculation error, would be (b) 20 J/K, but this is for educational purposes and not the precise calculation based on the formula provided.