One litre of a gas at `10^@C` is heated till both its volume and pressure are tripled. Find the new temperature.

Explanation

To solve the problem of finding the new temperature of a gas when both its volume and pressure are tripled, we can use the ideal gas law, which states that:

\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- Initial volume \( V_1 = 1 \, \text{L} \)
- Initial temperature \( T_1 = 10^\circ C \)
- Convert \( T_1 \) to Kelvin:
\[ T_1 = 10 + 273 = 283 \, K \]

2. **Identify Final Conditions:**
- Since the volume and pressure are tripled:
- Final volume \( V_2 = 3V_1 = 3 \times 1 = 3 \, \text{L} \)
- Final pressure \( P_2 = 3P_1 \)

3. **Set Up the Ideal Gas Law Equation:**
- Substitute the known values into the ideal gas law:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
- Substitute \( P_2 \) and \( V_2 \):
\[ \frac{P_1 \cdot 1}{283} = \frac{(3P_1) \cdot (3)}{T_2} \]

4. **Simplify the Equation:**
- Cancel \( P_1 \) from both sides:
\[ \frac{1}{283} = \frac{9}{T_2} \]

5. **Cross-Multiply to Solve for \( T_2 \):**
- Rearranging gives:
\[ T_2 = 9 \times 283 \]

6. **Calculate \( T_2 \):**
- Perform the multiplication:
\[ T_2 = 2547 \, K \]

### Final Answer:
The new temperature \( T_2 \) is \( 2547 \, K \).