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 between pressure and temperature for a gas at constant volume, which is described by Gay-Lussac's Law. This law states that the pressure of a gas is directly proportional to its absolute temperature when the volume is held constant.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial pressure (P1) = 4 atm
- Initial temperature (T1) = 27 °C
- Final temperature (T2) = 57 °C

2. **Convert Temperatures to Kelvin:**
- To convert Celsius to Kelvin, use the formula:
\[ T(K) = T(°C) + 273 \]
- For T1:
\[ T1 = 27 + 273 = 300 \, K \]
- For T2:
\[ T2 = 57 + 273 = 330 \, K \]

3. **Apply Gay-Lussac's Law:**
- According to Gay-Lussac's Law:
\[ \frac{P1}{T1} = \frac{P2}{T2} \]
- Rearranging the formula to find P2:
\[ P2 = P1 \times \frac{T2}{T1} \]

4. **Substitute the Known Values:**
- Substitute P1, T1, and T2 into the equation:
\[ P2 = 4 \, atm \times \frac{330 \, K}{300 \, K} \]

5. **Calculate P2:**
- Perform the calculation:
\[ P2 = 4 \times \frac{330}{300} \]
\[ P2 = 4 \times 1.1 = 4.4 \, atm \]

6. **Final Answer:**
- The pressure of the gas during driving (P2) is approximately **4.4 atm**.