Explanation:
To solve this problem, we can use the combined gas law, which relates the pressure, volume, and temperature of a gas:

\[
\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}
\]

Where:
- \(P_1\) and \(V_1\) are the initial pressure and volume
- \(T_1\) and \(T_2\) are the initial and final temperatures (in Kelvin)
- \(P_2\) and \(V_2\) are the final pressure and volume

Step-by-Step Solution:

1. Convert temperatures to Kelvin:
\[
T_1 = 0^\circ C = 273 \, \text{K} \quad \text{(S.T.P. temperature)}
\]
\[
T_2 = 91^\circ C = 91 + 273 = 364 \, \text{K}
\]

2. Initial conditions:
- \(P_1 = 1 \, \text{atm} = 760 \, \text{mm Hg}\) (S.T.P. pressure)
- \(V_1 = 30.0 \, \text{dm}^3\)
- \(T_1 = 273 \, \text{K}\)

3. Final conditions:
- \(P_2 = 380 \, \text{mm Hg}\)
- \(T_2 = 364 \, \text{K}\)
- We need to find \(V_2\).

4. Use the combined gas law:
\[
\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}
\]
Substitute the known values:
\[
\frac{760 \times 30.0}{273} = \frac{380 \times V_2}{364}
\]

5. Solve for \(V_2\):
\[
V_2 = \frac{760 \times 30.0 \times 364}{273 \times 380}
\]
\[
V_2 = \frac{8310720}{103740}
\]
\[
V_2 \approx 80.0 \, \text{dm}^3
\]

Thus, the volume of the gas at 91°C and 380 mm Hg would be 80.0 dm³, so the correct answer is D: 80.0 dm³.