A 10 litre gas is inserted into a car tyre at 4 atm and `27^@C`. The temperature of tyre increase to `57^@C` during driving. What would be the pressure of gas during driving?

Options

A. 3.1 atm
B. 4.4 atm
C. 300mmHg
D. 2270 mmHg

Answer

To solve the problem, we will use the relationship defined by Gay-Lussac's law, which states that the pressure of a gas is directly proportional to its absolute temperature when the volume is constant.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial pressure, \( P_1 = 4 \, \text{atm} \)
- Initial temperature, \( T_1 = 27^\circ C = 27 + 273 = 300 \, \text{K} \)
- Final temperature, \( T_2 = 57^\circ C = 57 + 273 = 330 \, \text{K} \)
- Volume, \( V = 10 \, \text{liters} \) (constant)

2. **Apply Gay-Lussac's Law:**
- According to Gay-Lussac's law:
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]
- Rearranging the equation to find \( P_2 \):
\[ P_2 = P_1 \times \frac{T_2}{T_1} \]

3. **Substitute the Known Values:**
- Substitute \( P_1 \), \( T_1 \), and \( T_2 \) into the equation:
\[ P_2 = 4 \, \text{atm} \times \frac{330 \, \text{K}}{300 \, \text{K}} \]

4. **Calculate \( P_2 \):**
- Perform the calculation:
\[ P_2 = 4 \times \frac{330}{300} = 4 \times 1.1 = 4.4 \, \text{atm} \]

5. **Final Answer:**
- The pressure of the gas during driving is \( P_2 = 4.4 \, \text{atm} \).