A gas at 50°C and 30 cm³ is heated to 80°C at constant pressure. What is the fractional increase in volume?

The initial temperature $T_1 = 50^\circ\text{C}$.
The final temperature $T_2 = 80^\circ\text{C}$.
To convert Celsius to Kelvin, add 273.15 (or 273 for common approximations). We will use 273.15 for better accuracy.

$$T_1 = 50^\circ\text{C} + 273.15 = 323.15 \text{ K}$$ $$T_2 = 80^\circ\text{C} + 273.15 = 353.15 \text{ K}$$

Charles's Law states that for a fixed amount of gas at constant pressure, the volume is directly proportional to its absolute temperature. The formula is:

$$\frac{V_1}{T_1} = \frac{V_2}{T_2}$$

Where $V_1$ is the initial volume, $T_1$ is the initial absolute temperature, $V_2$ is the final volume, and $T_2$ is the final absolute temperature.
We are given $V_1 = 30 \text{ cm}^3$. We need to find the fractional increase in volume, which is $\frac{V_2 - V_1}{V_1}$.

From Charles's Law, we can write:

$$\frac{V_2}{V_1} = \frac{T_2}{T_1}$$

Substitute the Kelvin temperatures:

$$\frac{V_2}{V_1} = \frac{353.15 \text{ K}}{323.15 \text{ K}}$$ $$\frac{V_2}{V_1} \approx 1.09279$$

The fractional increase in volume is defined as $\frac{\text{change in volume}}{\text{original volume}}$, which is $\frac{V_2 - V_1}{V_1}$.
This can be rewritten as $\frac{V_2}{V_1} - \frac{V_1}{V_1} = \frac{V_2}{V_1} - 1$.
Using the ratio calculated in Step 3:

$$\text{Fractional increase} = \frac{V_2}{V_1} - 1$$ $$\text{Fractional increase} = 1.09279 - 1$$ $$\text{Fractional increase} = 0.09279$$

Rounding 0.09279 to three decimal places gives 0.093.
Comparing this to the given options:
A) 0.009
B) 0.093
C) 0.910
D) 1.090

The calculated fractional increase matches option (B).

The final answer is $\boxed{\text{B}}$