Answer
b) $\omega = \frac{-W_{1-2}}{W_{2-3}}= \frac{1}{n-1}\frac{1-r^{1-n}}{r-1}$
c) $\omega = 0.2558$
Work Step by Step
a) Process 1-2 is defined by the hyperbolic equation $PV^n=c$ in the P-V diagram, whereas process 2-3 is a straight horizontal line.
b)For the polytropic process:
$W_{1-2} = \frac{P_1V_1-P_2V_2}{n-1} $
$W_{1-2} = (\frac{P_2V_2}{n-1}) (\frac{P_1V_1}{P_2V_2}-1)$
With: $P_1V_1^n=P_2V_2^n \Rightarrow \frac{P_1}{P_2} = r^{-n} $
$W_{1-2} = (\frac{P_2V_2}{n-1}) (r^{1-n}-1)$
For the isobaric process:
$W_{2-3} = P_2(V_3-V_2)$, With $V_3=V_1$
$W_{2-3} = P_2V_2(r-1)$
The rate $\omega = \frac{-W_{1-2}}{W_{2-3}}$
$\omega = \frac{1}{n-1}\frac{1-r^{1-n}}{r-1}$
c) With n=1.4 and r=6:
$\omega = 0.2558$
