An automobile tire is filled to a gauge pressure of `200\ kPa` at `10°C`. After a drive of `100\ km`, the temperature within the tire rises to `40°C`. What is the pressure within the tire now? Given: `T_1 = 10°C = 283 K` `P_1 = 200 kPa + 101\ kPa = 301\ kPa` `T_2 = 40°C = 313 K` `P_2 = ?` Equation: `PV = nRT` Solution: For fixed amount of gas, `nR` is a constant. Thus, `(PV)/T=` constant. or `(P_1V_1)/T_1 = (P_2V_2)/T_2` In this problem, we consider the volume inside the tire will not change. then `(P_1V)/T_1 = (P_2V)/T_2 => P_1/T_1 = P_2/T_2 ` `P_2 = P_1T_2/T_2` `P_2 = (301\ kPa)(313 K)/(283 K)` The gauge pressure will be