The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of 10 dm^3 to a volume of 100 dm^3 at 27°C is: A. 42.3 J/K·mol
B. 38.3 J/K·mol
C. 35.8 J/K·mol
D. 32.3 J/K·mol

To solve this problem, we need to use the formula for the entropy change (\(\Delta S\)) for the isothermal expansion of an ideal gas. The formula for entropy change during an isothermal process is:

\[
\Delta S = nR \ln \frac{V_f}{V_i}
\]

where:
- \(n\) is the number of moles of the gas,
- \(R\) is the universal gas constant,
- \(V_f\) is the final volume,
- \(V_i\) is the initial volume.

Given:
- Number of moles (\(n\)) = 2 moles
- Initial volume (\(V_i\)) = 10 \(\text{m}^3\)
- Final volume (\(V_f\)) = 100 \(\text{m}^3\)
- \(R\) = 8.314 J/(K·mol) (universal gas constant)

Plug these values into the formula:

\[
\Delta S = 2 \times 8.314 \ln \frac{100}{10}
\]

First, calculate the ratio:

\[
\frac{V_f}{V_i} = \frac{100}{10} = 10
\]

Now, find the natural logarithm of 10:

\[
\ln 10 \approx 2.302
\]

Next, multiply by \(nR\):

\[
\Delta S = 2 \times 8.314 \times 2.302
\]

Calculate:

\[
\Delta S \approx 2 \times 8.314 \times 2.302 \approx 38.3 \text{ J/K·mol}
\]

So the entropy change involved is approximately \(38.3 \text{ J/K·mol}\). Thus, the correct answer is:

**B. 38.3 J/K·mol**