10 g of a gas at 1 atm and 273 K occupies 5 litres. The temperature at which the volume becomes double for the same mass of gas at the same pressure is:

Options

A. 273 K
B. `-273^(@)C`
C. `273^(@)C`
D. `546^(@)C`

Explanation

To solve the problem, we will use the ideal gas law, which states that for a given amount of gas at constant pressure, the volume of the gas is directly proportional to its temperature in Kelvin.

### Step-by-Step Solution:

1. **Identify the known values:**
- Mass of gas (m) = 10 g
- Initial pressure (P1) = 1 atm
- Initial temperature (T1) = 273 K
- Initial volume (V1) = 5 L

2. **Determine the final volume (V2):**
- The problem states that the volume becomes double. Therefore:
\[ V2 = 2 \times V1 = 2 \times 5 \, \text{L} = 10 \, \text{L} \]

3. **Use the relationship between volume and temperature:**
- According to Charles's Law, at constant pressure, the ratio of the volumes and temperatures is given by:
\[ \frac{V1}{T1} = \frac{V2}{T2} \]
- Rearranging this gives:
\[ T2 = \frac{V2 \times T1}{V1} \]

4. **Substitute the known values into the equation:**
- Substitute \(V2 = 10 \, \text{L}\), \(T1 = 273 \, \text{K}\), and \(V1 = 5 \, \text{L}\):
\[ T2 = \frac{10 \, \text{L} \times 273 \, \text{K}}{5 \, \text{L}} = \frac{2730}{5} = 546 \, \text{K} \]

5. **Convert the final temperature from Kelvin to Celsius:**
- To convert Kelvin to Celsius, subtract 273:
\[ T2 \, (\text{in °C}) = T2 \, (\text{in K}) - 273 = 546 - 273 = 273 \, \text{°C} \]

6. **Conclusion:**
- The temperature at which the volume becomes double for the same mass of gas at the same pressure is **273 °C**.