0.2m^(3) of an ideal gas at a pressure of 2 MPa and 600 K is expanded isothermally to 5 times the initial volume. It is then cooled to 300 K at constant volume and then compressed back polytropically to its initial state. Determine the net work done and heat transfer during the cycle.

0.2m^(3) of an ideal gas at a pressure of 2 MPa and 600 K is expanded

1.	0.2m^(3) of an ideal gas at a pressure of 2 MPa and 600 K is expanded isothermally to 5 times the initial volume. It is then cooled to 300 K at constant volume and then compressed back polytropically to its initial state. Determine the net work done and heat transfer during the cycle.
Answer» Solution :Considering PROCESS `1-2` which is isothermal, i.e., `T_(1)=T_(2)=600K` `P_(2)=(p_(1)V_(1))/(V_(2)) = (2xx10^(6)xx0.2)/((5xx0.2))=400xx10^(3)N//m^(2)=0.4MPa` Work done, `W_(1-2)=p_(1)V_(1) "log"_(e ) (V_(2))/(V_(1)) =2xx10^(6)xx0.2log_(e )5` `=643775J=643.775kJ` Considering process 2-3 which is at constant volume, i.e, `V_(2) = V_(3) = 5 xx 0.2 = 1m^(3)` `P_(3)= p_(2)xx (T_(3))/(T_(2)) = 400 xx 10^(3) xx (300)/(600)` `=200xx10^(3)N//m^(2) =0.2 MPa` `W_(2-3)=0` as volume remains constant. For the polytropic process `3-1=p_(3) V_(3)^(n) =p_(1)V_(1)^(n)` `n=("log"_(e )(p_(1))/(p_(3)))/("log"_(e )(V_(3))/(V_(1)))=("log"_(e )(2)/(0.2))/("log"_(e )(1)/(0.2))=(2.3026)/(1.6094)=1.433` `W_(3-1)=(p_(3)V_(3)-p_(1)V_(1))/(n-1) =(0.2xx10^(6) xx1-2xx10^(6)xx0.2)/(1.433-1)` `= -461894J= -461.894kJ` New work done, `W_(n)=W_(1-2) +W_(2-3) +W_(3-1)` `=643.774+0+(-461.894)=181.88kJ` `:.` Heat TRANSFER during the complete cycle = 181.88 kJ