At 1 atmospheric pressure and `0 .^(@)C` , certain mass of a gas measures `0.4 L` . Keeping the pressure constant , if the temperature is increased is increased to `273 .^(@)C` , what will be its volume ?

Options

A. 0.8 L
B. 22.4 L
C. 54.6 L
D. 0.4 L

Explanation

To solve the problem, we will use Charles's Law, which states that the volume of a gas is directly proportional to its temperature (in Kelvin) when pressure is held constant. The formula we will use 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. **Identify the given values:**
- Initial volume \( V_1 = 0.4 \, \text{L} \)
- Initial temperature \( T_1 = 0^\circ C = 273 \, \text{K} \) (convert Celsius to Kelvin by adding 273)
- Final temperature \( T_2 = 273^\circ C = 546 \, \text{K} \) (convert Celsius to Kelvin by adding 273)

2. **Set up the equation using Charles's Law:**
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]

3. **Rearrange the equation to solve for \( V_2 \):**
\[ V_2 = V_1 \times \frac{T_2}{T_1} \]

4. **Substitute the known values into the equation:**
\[ V_2 = 0.4 \, \text{L} \times \frac{546 \, \text{K}}{273 \, \text{K}} \]

5. **Calculate the ratio:**
\[ \frac{546}{273} = 2 \]

6. **Calculate \( V_2 \):**
\[ V_2 = 0.4 \, \text{L} \times 2 = 0.8 \, \text{L} \]

### Final Answer:
The final volume \( V_2 \) is \( 0.8 \, \text{L} \).

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