Answer:
Explanation:
Given that:
From process 1 → 2
[tex]P_1 = 10 bar \\ \\ V_1 = 1 m^3 \\ \\ V_2 = 4 m^3[/tex]
[tex]PV^{1.5} = \ constant[/tex]
[tex]\gamma = 1.5[/tex]
Process 2 → 3
The volume is constant i.e [tex]V_2 =V_3 = 4m^3[/tex]
[tex]P_3 = 10 \ bar[/tex]
Process 3 → 1
P = constant i.e the compression from state 1
Now, to start with 1 → 2
[tex]P_1V_1^{1.5} = P_2V_2^{1.5}[/tex]
[tex]P_2 = P_1 (\dfrac{V_1}{V_2})^{1.5}[/tex]
[tex]P_2 = 10 \times (\dfrac{1}{4})^{1.5}[/tex]
[tex]P_2 =1.25[/tex]
The work-done for the process 1 → 2 through adiabatic expansion is:
[tex]W = \dfrac{1}{1-\gamma}[P_2V_2-P_1V_1][/tex]
We know that 1 bar = [tex]10^5 \ N/m^2[/tex]
∴
[tex]W = \dfrac{1}{1-1.5}[1.25 \times 10^5 \times 4- 10 \times 10^5 \times 1][/tex]
[tex]W =1000000 \ J[/tex]
[tex]W_{1 \to 2} = 1000 kJ[/tex]
For process 2 → 3
Since V is constant
Thus:
W = PΔV = 0
[tex]W_{2 \to 3} = 0[/tex]
For process 3 → 1
W = PΔV
[tex]W _{3 \to 1} = P_3(V_1-V_3)[/tex]
[tex]W _{3 \to 1} = 10 \times 10^5 (1-4)[/tex]
[tex]W _{3 \to 1} = 10 \times 10^5 (-3)[/tex]
[tex]W _{3 \to 1} = -3 \times 10^6 \ J[/tex]
[tex]W _{3 \to 1} = -3000 \ kJ[/tex]
The net work-done now for the entire system is :
[tex]W_{net} = W_{1 \to 2} + W_{2 \to 3 } + W_{ 3 \to 1 }[/tex]
[tex]W_{net} = (1000 + 0 + (-3000)) \ kJ[/tex]
[tex]W_{net} =-2000 \ kJ[/tex]
The sketch of the processes on p -V coordinates can be found in the image attached below.
