1 L of air weighs 1.293 g at `0^(@)C` and 1 atm pressure. At becomes `30 dm^(3)` . To what temperature the gas must be raised to accomplish the change ?

Explanation

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

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

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Initial volume, \( V_1 = 1 \, \text{L} = 1 \, \text{dm}^3 \)
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