Answer:
Explanation:
For first isobaric process (expansion with constant pressure):
[tex]W_{1}=P_{1}(V_{2}-V_{1})=(0.43\times 10^{6})(2.9-0.28)=1.1266 \times 10^{6} J=1.1266 MJ(+) (Expansion)[/tex]
Now, for the first isochoric: [tex]W_{2}=0[/tex], and [tex]Q_{2}=\Delta U=U_{3}-U_{2}[/tex]
For the second isobaric process (compression with constant pressure):
[tex]W_{3}=P_{2}(V_{1}-V_{2})=(4.3\times 10^{6})(0.28-2.9) =-11.266\times 10^{6}J=-11.266 MJ (Compression)[/tex]
For the last isochoric: [tex]W_{4}=0, Q_{4}=\Delta U=U_{4}-U_{3}[/tex]
So, the total work per cycle: [tex]W=W_{1}+W_{2}+W_{3}+W_{4}=1.1266+0-11.266+0=-10.1394 MJ[/tex]
