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?

Introduction / Context:
Entropy change during isothermal reversible processes is a key topic in thermodynamics. For an ideal gas undergoing an isothermal reversible expansion, the entropy change depends on the number of moles, the gas constant and the volume ratio. This question tests whether you can recall and correctly apply the formula for entropy change in such a process. It also checks your ability to handle natural logarithms and interpret units in J/mol/K for a multi mole system.

Given Data / Assumptions:

• Number of moles, n = 2 mol of ideal gas.
• Initial volume, V1 = 10 dm3.
• Final volume, V2 = 100 dm3.
• Temperature is isothermal at 27°C, which is 300 K, but the formula for entropy change in volume based isothermal expansion does not require T directly.
• The process is reversible and the gas behaves ideally.

Concept / Approach:
For an ideal gas undergoing an isothermal reversible change, the entropy change of the gas can be written in terms of volume as ΔS = n * R * ln(V2 / V1). Here, n is the number of moles, R is the gas constant in J/mol/K, and ln denotes the natural logarithm. The temperature is constant, so this formula is appropriate. Our task is to compute the volume ratio V2 / V1, take the natural logarithm, multiply by n and R, and then match the result with the closest option given in the question.

Step-by-Step Solution:
Step 1: Write the formula for entropy change for isothermal reversible expansion of an ideal gas: ΔS = n * R * ln(V2 / V1).Step 2: Substitute the known values: n = 2 mol, V1 = 10 dm3 and V2 = 100 dm3.Step 3: Compute the volume ratio: V2 / V1 = 100 / 10 = 10.Step 4: Use R ≈ 8.314 J/mol/K.Step 5: Calculate ln(10). The natural logarithm of 10 is approximately 2.3026.Step 6: Multiply the terms: ΔS = 2 * 8.314 * 2.3026.Step 7: First compute 8.314 * 2.3026 ≈ 19.148.Step 8: Then multiply by 2: 2 * 19.148 ≈ 38.296 J/mol/K.Step 9: Round the result to one decimal place, giving ΔS ≈ 38.3 J/mol/K.

Verification / Alternative check:
As a quick check, you can estimate ln(10) as about 2.3 and R as about 8.3. Then ΔS ≈ 2 * 8.3 * 2.3. Multiplying 8.3 by 2.3 gives approximately 19.1, and multiplying this by 2 yields about 38.2. This rough estimate is very close to 38.3 J/mol/K, supporting the more precise calculation. None of the other options are as close to this value, confirming that 38.3 J/mol/K is the appropriate choice from the list.

Why Other Options Are Wrong:
Option a, 42.3 J/mol/K, would require a larger logarithm or different values of n or R, which are not consistent with the data. Options c, 35.8 J/mol/K, and d, 32.3 J/mol/K, are significantly lower than the calculated result and would correspond to smaller volume ratios or fewer moles of gas. Since we know the volume ratio is exactly 10 and the number of moles is 2, these values cannot be correct. Only 38.3 J/mol/K matches the formula based calculation for the given conditions.

Common Pitfalls:
One common mistake is to forget to use the natural logarithm and instead use base 10 logarithms, which leads to an incorrect value of ln(V2 / V1). Another error is to misinterpret the volume ratio, using V1 / V2 instead of V2 / V1, which would give a negative entropy change, contradicting the fact that entropy increases during expansion. Students may also forget to multiply by the number of moles, using n = 1 instead of n = 2. Carefully writing the formula and substituting each value step by step helps avoid these issues.

Final Answer:
The entropy change for the isothermal reversible expansion described is 38.3 J/mol/K.