An ideal gas initially at $600K$ and $10$ bar undergoes a four$-$step mechanically

reversible cycle in a closed system. In step $1-2,$ pressure decreases isothermally to $3$

bar; in step $2-3,$ pressure decrease at constant volume to $\frac{2}{b}ar$ ; in step $3-4,$ volume

decreases at constant pressure; and in step $4-1,$ the gas returns adiabatically to its initial

state. Take $C_p=(\frac{7}{2})R$ and $C_v=(\frac{5}{2})R.$

$($a$)$ Sketch the cycle on a PV diagram.

$($b$)$ Determine T$,$ P$,$ V for states $1,2,3,$ and $4$ where unknown.

$($c$)$ Calculate q$,$ w$,U$ and $H$ for each step of the cycle.

Answer:

$($a$)$ Sketch the cycle on a PV diagram $($Answer omitted$)$

$($b$)$ Determine $($where unknown$)$ both $T$ and $P$ for states $1,2,3,$ and $4.$

Point $1$: $P_1=10$ bar $,T_1=600K,V_1=4\mathrm{.}988{10}^{-3}\frac{m^3}{m}ol$

Point $2$: $P_2=\frac{3}{b}ar,T_2=600K,V_2=0\mathrm{.}017\frac{m^3}{m}ol$

Point $3$: $P_3=2$ bar, $T_3=409K,V_3=0\mathrm{.}017\frac{m^3}{m}ol$

Point $4$: $P_4=2$ bar, $T_4=378\mathrm{.}831K,V_4=0\mathrm{.}016\frac{m^3}{m}ol$

$($c$)$ Calculate $Q,W,U$ and $H$ for each step of the cycle.

Step$12$: $U_{12}=0\frac{J}{m}ol,H_{12}=0\frac{J}{m}ol,w_{12}=-6006\frac{J}{m}ol,-q_{12}=-6006\frac{J}{m}ol$

Step$23$: $U_{23}=-4\mathrm{.}157{10}^3\frac{J}{m}ol,H_{23}=-5\mathrm{.}82{10}^3\frac{J}{m}ol,$

$w_{23}=0\frac{J}{m}ol,q_{23}=-4\mathrm{.}157{10}^3\frac{J}{m}ol$

Step$34$: $U_{34}=-627\frac{J}{m}ol,H_{34}=-877\mathrm{.}9\frac{J}{m}ol$

$q_{34}=-877\mathrm{.}9\frac{J}{m}ol,w_{34}=250\mathrm{.}8\frac{J}{m}ol$

Step$41$: $U_{41}=4\mathrm{.}597{10}^3\frac{J}{m}ol,H_{41}=6\mathrm{.}436{10}^3\frac{J}{m}ol$