Answer:
Required compressor work W=8560.44 KJ/Kmol
Explanation:
Given that
Initial pressure = 1 bar
[tex]P_1=1\ bar[/tex]
Final pressure = 40 bar
[tex]P_2=40\ bar[/tex]
Process is isothermal and T=280 K
We know that ,work done in isothermal process given as
[tex]W=P_1V_1\ln \dfrac{P_1}{P_2}[/tex]
given taht gas is ideal so
P V =m R T
[tex]W=mRT\ln \dfrac{P_1}{P_2}[/tex]
R for ethylene
R=0.296 KJ/kg.K
Now by putting the values
[tex]W=mRT\ln \dfrac{P_1}{P_2}[/tex]
[tex]W=1\times 0.296\times 280 \ln \dfrac{1}{40}[/tex]
W= -305.73 KJ/kg
Negative sign indicates that work done on the system.
Required compressor work W=305.73 KJ/kg
Molar mass of ethylene M= 28 Kg/Kmol
So W= 305.73 x 28 KJ/Kmol
W=8560.44 KJ/Kmol
Required compressor work W=8560.44 KJ/Kmol
