Answer
a)$ P_f = 571 \ kPa$
b) $-677.74 \ kJ$
Work Step by Step
a) We know that for adiabatic processes, $PV^{\gamma}$ is constant. Thus, we find:
$(60)(5^{1.4})=P_f(1^{1.4}) \\ P_f = 571 \ kPa$
b) We use the equation for work for an adiabatic process, which is equation 18.12 in the textbook. Doing this, we find:
$W = \frac{(571.096)(1)-(60)(5)}{1.4-1}=-677.74 \ kJ$
