A car tyre contains air at a gauge pressure of 2 atm and a...

The initial absolute pressure of the air in the tyre is $P_1 = P_{\text{gauge}} + P_{\text{atm}} = 2 + 1 = 3 \text{ atm}$.

From the ideal gas equation, $PV = nRT$, the absolute pressure is directly proportional to the product of the number of moles and the absolute temperature for a constant volume ($P \propto nT$).

Since the mass of the air increases by $20\%$, the new number of moles is $n_2 = 1.2 n_1$. The initial temperature is $T_1 = 300 \text{ K}$ and the final temperature is $T_2 = 330 \text{ K}$.

The final absolute pressure $P_2$ is given by:
$P_2 = P_1 \times \left(\dfrac{n_2}{n_1}\right) \times \left(\dfrac{T_2}{T_1}\right)$
$P_2 = 3 \times 1.2 \times \left(\dfrac{330}{300}\right) = 3 \times 1.2 \times 1.1 = 3.96 \text{ atm}$.

The final gauge pressure is $P_{2,\text{gauge}} = P_2 - P_{\text{atm}} = 3.96 - 1 = 2.96 \text{ atm}$.

Answer: $2.96 \text{ atm}$