Step 1: Understanding the Concept:
For a gas enclosed in a rigid container like a vehicle tyre (assuming the volume expansion of the tyre is negligible), the process is isochoric (constant volume).
According to Gay-Lussac's Law, for a fixed mass and constant volume of an ideal gas, the pressure is directly proportional to its absolute temperature (in Kelvin).
Step 2: Key Formula or Approach:
Gay-Lussac's Law equation:
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]
where $P_1, T_1$ are the initial pressure and absolute temperature, and $P_2, T_2$ are the final pressure and absolute temperature.
Note: Temperatures MUST be converted to Kelvin. $T(K) = T(^\circ\text{C}) + 273$.
Step 3: Detailed Explanation:
Given values:
Initial pressure, $P_1 = 270\text{ kPa}$
Initial temperature, $t_1 = 27^\circ\text{C} \implies T_1 = 27 + 273 = 300\text{ K}$
Final temperature, $t_2 = 37^\circ\text{C} \implies T_2 = 37 + 273 = 310\text{ K}$
We need to find final pressure $P_2$.
Rearranging the formula to solve for $P_2$:
\[ P_2 = P_1 \times \left( \frac{T_2}{T_1} \right) \]
Substitute the values:
\[ P_2 = 270 \times \left( \frac{310}{300} \right) \]
\[ P_2 = 270 \times \left( \frac{31}{30} \right) \]
\[ P_2 = \left( \frac{270}{30} \right) \times 31 \]
\[ P_2 = 9 \times 31 \]
\[ P_2 = 279\text{ kPa} \]
Step 4: Final Answer:
The new air pressure in the tyre is $279\text{ kPa}$.