A flask of capacity 10 litre containing air is heated from `27^(@)C " to " 327^(@)C`. The ratio of mole of air present at `27^(@)C` to mole present at `327^(@)C` is ____________.

Explanation

To solve the problem, we will use the Ideal Gas Law, which states that \( PV = nRT \). Here, \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial temperature \( T_1 = 27^\circ C \)
- Final temperature \( T_2 = 327^\circ C \)
- Volume \( V = 10 \, \text{liters} \) (constant)
- Pressure \( P \) (constant)

2. **Convert Temperatures to Kelvin:**
- To convert Celsius to Kelvin, use the formula:
\[ T(K) = T(°C) + 273 \]
- For \( T_1 \):
\[ T_1 = 27 + 273 = 300 \, K \]
- For \( T_2 \):
\[ T_2 = 327 + 273 = 600 \, K \]

3. **Use the Ideal Gas Law Relationship:**
- Since the pressure and volume are constant, we can use the relationship:
\[ n_1 T_1 = n_2 T_2 \]
- Rearranging gives us:
\[ \frac{n_1}{n_2} = \frac{T_2}{T_1} \]

4. **Substitute the Values:**
- Substitute \( T_1 \) and \( T_2 \) into the equation:
\[ \frac{n_1}{n_2} = \frac{600}{300} \]

5. **Calculate the Ratio:**
- Simplifying gives:
\[ \frac{n_1}{n_2} = 2 \]
- Therefore, the ratio of moles of air present at \( 27^\circ C \) to moles present at \( 327^\circ C \) is:
\[ \frac{n_1}{n_2} = 2:1 \]

### Final Answer:
The ratio of moles of air present at \( 27^\circ C \) to moles present at \( 327^\circ C \) is \( 2:1 \).