Question

If a tire has a pressure of 11.9 Pa, a volume of 2.8 L at a temperature of 319 K, and the pressure increases to 13.4 kPa when the temperature rises to 325 K, what is the new volume of the tire?

Answer 1

Approx. `2.5*L`

Explanation:

We use the combined gas law, `(P_1V_1)/T_1=(P_2V_2)/T_2`

Thus `V_2=(P_1*T_2*V_1)/(P_2*T_1)` `=` `(11.9*Paxx325*Kxx2.8*L)/(13.4*Paxx319*K)`

So pressure seems to win over temperature here inasmuch as the starting volume has been reduced.