Answer
a) $172.99 \ kPa$
b) $268.685 \ J$
Work Step by Step
a) We know that for adiabatic processes, $PV^{\gamma}$ is constant. Thus, we find:
$(98.5)(6.25^{1.4})=P_f(4.18^{1.4}) \\ P_f = 172.99 \ kPa$
b) We use the equation for work for an adiabatic process, which is equation 18.12 in the textbook. Doing this, we find:
$W = \frac{(172.99)(4.18)-(98.5)(6.25)}{1.4-1}=268.685 \ J$
