A perfect gas at `27^(@)C` is heated at constant pressure so as to triple its volume. The tmemperature of th gas will be

Options

A. `84^(@)C`
B. `900^(@)C`
C. `627^(@)C`
D. `450^(@)C`

Explanation

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

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- Initial temperature \( T_1 = 27^\circ C \)
- Convert \( T_1 \) to Kelvin:
\[ T_1 = 27 + 273 = 300 \, K \]

2. **Identify the Final Conditions**:
- The volume of the gas is tripled, so:
\[ V_2 = 3V_1 \]

3. **Apply Charles's Law**:
- According to Charles's Law:
\[ \frac{V_1}{V_2} = \frac{T_1}{T_2} \]
- Substituting the known values:
\[ \frac{V_1}{3V_1} = \frac{300 \, K}{T_2} \]
- This simplifies to:
\[ \frac{1}{3} = \frac{300}{T_2} \]

4. **Solve for \( T_2 \)**:
- Cross-multiply to solve for \( T_2 \):
\[ T_2 = 300 \times 3 = 900 \, K \]

5. **Convert \( T_2 \) Back to Celsius**:
- Convert \( T_2 \) from Kelvin to Celsius:
\[ T_2 = 900 - 273 = 627^\circ C \]

### Final Answer:
The temperature of the gas when its volume is tripled at constant pressure is \( 627^\circ C \).

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