At what temperature volume of an ideal gas at `0^@C` becomes triple

Options

A. `546^@C`
B. `182^@C`
C. `819^@C`
D. `646^@C`

Explanation

To solve the problem of finding the temperature at which the volume of an ideal gas at \(0^\circ C\) becomes triple, we can use the ideal gas law and the relationship between volume and temperature under constant pressure.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- Initial temperature, \(T_1 = 0^\circ C\).
- Initial volume, \(V_1 = V\).
- Final volume, \(V_2 = 3V\).

2. **Convert Initial Temperature to Kelvin**:
- The temperature in Kelvin is given by the formula:
\[ T(K) = T(°C) + 273 \]
- Therefore,
\[ T_1 = 0 + 273 = 273 \, K \]

3. **Use the Relationship Between Volume and Temperature**:
- For an ideal gas at constant pressure, the volume is directly proportional to the temperature:
\[ \frac{T_1}{T_2} = \frac{V_1}{V_2} \]
- Substituting the known values:
\[ \frac{273}{T_2} = \frac{V}{3V} \]
- Simplifying gives:
\[ \frac{273}{T_2} = \frac{1}{3} \]

4. **Cross-Multiply to Solve for \(T_2\)**:
- Cross-multiplying gives:
\[ 273 \cdot 3 = T_2 \]
- Therefore,
\[ T_2 = 819 \, K \]

5. **Convert \(T_2\) Back to Celsius**:
- To convert Kelvin back to Celsius:
\[ T(°C) = T(K) - 273 \]
- Thus,
\[ T_2(°C) = 819 - 273 = 546 \, °C \]

### Final Answer:
The temperature at which the volume of the ideal gas becomes triple is \(546 \, °C\).

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