Problem 16.25

16.25. a)
Start with our favorite gas equation.
PV = nRT
Initially,
$(1 atm)(V_{0}) = nR(283{}^{\circ}K)$
and finally,
$(P_{f})(.28 V_{0}) = nR(313{}^{\circ}K)$
Dividing these two equations yields
$P_{f} = 3.95 atm ~=~ 4.00 \times 10^{5} Pa$
That is the absolute pressure. If you were to put a tire gauge on the tire, it would read this amount minus atmospheric pressure.

b)
After being driven,
$(P_{d})(1.02)(.28 V_{0}) = nR(358{}^{\circ}K)$
or P_d = 1.12 P_f
$P_{d} = 4.48 \times 10^{5} Pa ~=~ 50.4 lb/in^{2}$