A perfect gas at `27^(@)C` is heated at constant pressure to `327^(@)C`. If original volume of gas at `27^(@)C` is V, then volume at `327^(@)C` is

Options

A. V
B. `3V`
C. `2V`
D. `V//2`

Explanation

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

### Step-by-Step Solution:

1. **Convert Celsius 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 \( 327^\circ C \).
- To convert to Kelvin:
\[ T_2 = 327 + 273 = 600 \, K \]

2. **Apply Charles's Law**:
- According to Charles's Law, we have:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
- Rearranging gives:
\[ V_2 = V_1 \cdot \frac{T_2}{T_1} \]

3. **Substitute the values**:
- We know \( V_1 = V \), \( T_1 = 300 \, K \), and \( T_2 = 600 \, K \).
- Substituting these values into the equation:
\[ V_2 = V \cdot \frac{600}{300} \]
- Simplifying this gives:
\[ V_2 = V \cdot 2 = 2V \]

4. **Final Answer**:
- The volume of the gas at \( 327^\circ C \) is \( 2V \).