Answer:
a. [tex]W_{1-2}=0\ J[/tex]
b. [tex]W_{2-3}=-554517.74\ J[/tex]
c. [tex]W_{3-4}=-400\ kJ[/tex]
Explanation:
Given:
pressure at point 1, [tex]P_1=1\ bar=100000\ Pa[/tex]
pressure at point 2, [tex]P_2=2\ bar=200000\ Pa[/tex]
volume at point 1, [tex]V_1=V_2=4\ m^3[/tex]
volume at point 3, [tex]V_3=2\ m^3[/tex]
volume at point 2, [tex]V_2=4\ m^3[/tex]
pressure at point 2, [tex]P_2=2\ bar=200000\ Pa[/tex]
volume at point 4, [tex]V_4=1\ m^3[/tex]
a.
Work done in process 1-2:
Being isochoric process:
[tex]dV=0[/tex]
and
[tex]W=P.dV[/tex]
so,
[tex]W_{1-2}=0\ J[/tex]
b.
Now we apply the ideal gas equation for the process 2-3:
[tex]P_2.V_2=P_3.V_3[/tex]
[tex]200000\times 4=P_3\times 2[/tex]
[tex]P_3=400000\ Pa=4\ bar[/tex]
Now work done in an isothermal process:
[tex]W_{2-3}=P_2.V_2\times ln(\frac{V_3}{V_2} )[/tex]
[tex]W_{2-3}=200000\times 4\times\ln(\frac{2}{4} )[/tex]
[tex]W_{2-3}=-554517.74\ J[/tex] -ve means work is consumed by the system
c.
Now for the isobaric process 3-4 we apply the ideal gas law:
Since the process is isobaric so,
[tex]P_4=P_3=4\ bar[/tex]
[tex]V_3=2\ m^3[/tex]
[tex]V_4=1\ m^3[/tex]
Now the work done:
[tex]W_{3-4}=P_3.dV[/tex]
[tex]W_{3-4}=400000\times (1-2)[/tex]
[tex]W_{3-4}=-400\ kJ[/tex]
