One litre of helium gas at a pressure `76 cm`. Of Hg and temperature `27^(@)C` is heated till its pressure and volume are double. The final temperature attained by the gas is:

Options

A. `327^(@)C`
B. `927^(@)C`
C. `1027^(@)C`
D. `827^(@)C`

Explanation

To solve the problem, we will use the ideal gas law and the relationship between pressure, volume, and temperature given by Gay-Lussac's law.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial volume, \( V_1 = 1 \, \text{L} \)
- Initial pressure, \( P_1 = 76 \, \text{cm Hg} \)
- Initial temperature, \( T_1 = 27^\circ C \)

2. **Convert the Initial Temperature to Kelvin:**
\[ T_1 = 27 + 273 = 300 \, \text{K} \]

3. **Determine the Final Pressure and Volume:**
- The problem states that both the pressure and volume are doubled.
- Final volume, \( V_2 = 2 \times V_1 = 2 \times 1 = 2 \, \text{L} \)
- Final pressure, \( P_2 = 2 \times P_1 = 2 \times 76 = 152 \, \text{cm Hg} \)

4. **Apply Gay-Lussac's Law:**
According to Gay-Lussac's law, we have:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

5. **Substitute the Known Values into the Equation:**
\[ \frac{76 \times 1}{300} = \frac{152 \times 2}{T_2} \]

6. **Cross-Multiply to Solve for \( T_2 \):**
\[ 76 \times 1 \times T_2 = 152 \times 2 \times 300 \]
\[ 76 T_2 = 91200 \]

7. **Isolate \( T_2 \):**
\[ T_2 = \frac{91200}{76} \]
\[ T_2 = 1200 \, \text{K} \]

8. **Convert the Final Temperature to Celsius:**
\[ T_2 (\text{in } ^\circ C) = T_2 (\text{in K}) - 273 \]
\[ T_2 = 1200 - 273 = 927 \, ^\circ C \]

### Final Answer:
The final temperature attained by the gas is \( 927 \, ^\circ C \).