Answer
$Q=-6.51\ kJ$
Work Step by Step
For an ideal gas:
$PV=mRT$
Given $P_1=100\ kPa, V_1=0.2\ m³,\ R=2.0769\ kJ/kg.K,\ T_1=283\ K$,
and solving for the mass: $m=0.03403\ kg$
Given $P_2=700\ kPa,\ T_2=563\ K$,
and solving for the volume: $V_2=0.05684\ m³$
The polyt\ kJr
opic relation is:
$P_2V_2^n=P_1V_1^n\rightarrow \ln{\frac{P_2}{P_1}}=n\ln{\frac{V_1}{V_2}}$
hence $n=1.547$
The boundary work is given by:
$W_b=-\frac{mR\Delta T}{n-1}$
$W_b=-36.19\ kJ$
The energy balance for this process is:
$Q-W_b=\Delta U=mc_v\Delta T$
Given $c_v=3.1156\ kJ/kg$:
$Q=-6.51\ kJ$
