Answer
$T_2=693.33\ °R=233.33°F$
$W_s=-59.3\ Btu$
Work Step by Step
Considering the air an ideal gas, and since the system has constant volume:
$\frac{T_2}{T_1}=\frac{P_2}{P_1}$
Given $P_2=40\ psia, P_1=30\ psia, T_1=520\ °R$:
we get to: $T_2=693.33\ °R=233.33°F$
The energy balance for this system reduces to:
$-W_s=\Delta U=mc_v(T_2-T_1)$
With $m=2\ lbm,\ c_v=0.171\ Btu/lbm.°R$:
we have: $W_s=-59.3\ Btu$
