A gas is heated from `0^(@)C` to `100^(@)C` at 1.0 atm pressure. If the initial volume of the gas is 10.0L, its final volume would be:

Options

A. 7.32L
B. 10.00L
C. 13.66L
D. 20.00L

Answer

To solve the problem, we will use Charles's Law, which states that for a given mass of gas at constant pressure, the volume of the gas is directly proportional to its absolute temperature (in Kelvin). The formula for Charles's Law is:

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

Where:
- \( V_1 \) = initial volume
- \( T_1 \) = initial temperature in Kelvin
- \( V_2 \) = final volume
- \( T_2 \) = final temperature in Kelvin

### Step-by-Step Solution:

1. **Convert the temperatures from Celsius to Kelvin:**
- The initial temperature \( T_1 = 0^\circ C = 273 \, K \)
- The final temperature \( T_2 = 100^\circ C = 373 \, K \)

2. **Identify the initial volume:**
- The initial volume \( V_1 = 10.0 \, L \)

3. **Set up the equation using Charles's Law:**
- According to Charles's Law:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
Substituting the known values:
\[ \frac{10.0 \, L}{273 \, K} = \frac{V_2}{373 \, K} \]

4. **Cross-multiply to solve for \( V_2 \):**
\[ V_2 = \frac{10.0 \, L \times 373 \, K}{273 \, K} \]

5. **Calculate \( V_2 \):**
\[ V_2 = \frac{3730 \, L \cdot K}{273 \, K} \approx 13.66 \, L \]

### Final Answer:
The final volume \( V_2 \) of the gas when heated from \( 0^\circ C \) to \( 100^\circ C \) at a constant pressure of 1.0 atm is approximately **13.66 L**.