Answer:
29970 J
Explanation:
Given:
[tex]\begin{gathered} P_A=P_B=10^5Pa \\ P_c=P_D=100\text{ Pa\lparen The new lower pressure\rparen} \\ V_A=V_D=0.1m^3 \\ V_B=V_C=0.4m^3 \end{gathered}[/tex]
Work done across AB
[tex]\begin{gathered} W_{AB}=P_{AB}(V_B-V_A) \\ \\ W_{AB}=10^5(0.4-0.1) \\ \\ W_{AB}=0.3\times10^5 \\ \\ W_{AB}=30000J \end{gathered}[/tex]
Work done across BC = 0 J (Since there is no change in volume)
Work done across DA = 0 J (Since there is no change in volume)
Work done across CD
[tex]\begin{gathered} W_{CD}=P_{CD}(V_D-V_C) \\ \\ W_{CD}=100(0.1-0.4) \\ \\ W_{CD}=100(-0.3) \\ \\ W_{CD}=-30J \end{gathered}[/tex]
Work done in one cycle:
[tex]\begin{gathered} W_{cycle}=W_{AB}+W_{BC}+W_{DA}+W_{CD} \\ \\ W_{cycle}=3000J+0+0-30 \\ \\ W_{cycle}=29970J \end{gathered}[/tex]
29970J of work will be done in a single cycle
