`300 ml` of a gas at `27^(@)C` is cooled to `-3^(@)C` at constant pressure, the final volume is

Options

A. `540ml`
B. `135ml`
C. `270 ml`
D. `350 ml`

Explanation

To solve the problem of finding the final volume of a gas when it is cooled from \(27^\circ C\) to \(-3^\circ C\) at constant pressure, we can 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.

### Step-by-Step Solution:

1. **Convert Temperatures to Kelvin:**
- The initial temperature \(T_1\) is \(27^\circ C\).
- To convert to Kelvin:
\[ T_1 = 27 + 273 = 300 \, K \]
- The final temperature \(T_2\) is \(-3^\circ C\).
- To convert to Kelvin:
\[ T_2 = -3 + 273 = 270 \, K \]

2. **Identify Initial Volume:**
- The initial volume \(V_1\) is given as \(300 \, ml\).

3. **Use Charles's Law:**
- According to Charles's Law:
\[ \frac{V_1}{V_2} = \frac{T_1}{T_2} \]
- Rearranging this gives:
\[ V_2 = V_1 \cdot \frac{T_2}{T_1} \]

4. **Substitute the Known Values:**
- Substitute \(V_1 = 300 \, ml\), \(T_1 = 300 \, K\), and \(T_2 = 270 \, K\):
\[ V_2 = 300 \cdot \frac{270}{300} \]

5. **Calculate the Final Volume:**
- Simplifying the equation:
\[ V_2 = 300 \cdot 0.9 = 270 \, ml \]

6. **Conclusion:**
- The final volume \(V_2\) after cooling the gas to \(-3^\circ C\) at constant pressure is:
\[ \boxed{270 \, ml} \]