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, which states that for a given amount of gas, the ratio of pressure (P), volume (V), and temperature (T) remains constant. The relationship can be expressed as:

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

### Step 1: Convert the initial temperature to Kelvin
The initial temperature \( T_1 \) is given as \( 27^\circ C \). To convert Celsius to Kelvin, we use the formula:

\[ T(K) = T(°C) + 273.15 \]

So,

\[ T_1 = 27 + 273.15 = 300.15 \, K \approx 300 \, K \, (\text{for simplicity}) \]

### Step 2: Identify the initial conditions
From the problem, we have the following initial conditions:
- Volume \( V_1 = 1 \, L \)
- Pressure \( P_1 = 76 \, cm \, Hg \)

### Step 3: Identify the final conditions
The problem states that both the pressure and volume are doubled:
- Final volume \( V_2 = 2 \, L \)
- Final pressure \( P_2 = 2 \times 76 \, cm \, Hg = 152 \, cm \, Hg \)

### Step 4: Substitute values into the ideal gas law equation
Now, substituting the known values into the equation:

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

Substituting the values:

\[ \frac{76 \, cm \, Hg \times 1 \, L}{300 \, K} = \frac{152 \, cm \, Hg \times 2 \, L}{T_2} \]

### Step 5: Solve for \( T_2 \)
Cross-multiplying gives:

\[ 76 \, cm \, Hg \times 1 \, L \times T_2 = 152 \, cm \, Hg \times 2 \, L \times 300 \, K \]

This simplifies to:

\[ 76 T_2 = 91200 \]

Now, solving for \( T_2 \):

\[ T_2 = \frac{91200}{76} \approx 1200 \, K \]

### Step 6: Convert \( T_2 \) back to Celsius (if needed)
To convert back to Celsius:

\[ T(°C) = T(K) - 273.15 \]

So,

\[ T_2(°C) = 1200 - 273.15 \approx 926.85 \, °C \approx 927 \, °C \]

### Final Answer
The final temperature attained by the gas is approximately \( 927 \, °C \).

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