Answer:
The net work produced = -36737.52 J
The efficiency of the cycle = 72.2%
Explanation:
Given that :
The temperature of the hot reservoir [tex](T__H})[/tex] = 800 °C = (800+273)K
The temperature of the cold reservoir [tex](T__C})[/tex] = 25 °C (25+273)K
Pressure [tex](P__A})[/tex] = 0.2 bar
Pressure [tex](P_B})[/tex] = 60 bar
Rate constant (R) = 8.314
Determination of the efficiency of the cycle (η) is given by the formula:
(η) = [tex]1-\frac{T__C}{T__H}[/tex]
= 1 - [tex]\frac{(800+273)K}{(25+273)K}[/tex]
= 0.722
= 72.2 %
∴ The efficiency of the cycle = 72.2 %
However, the heat given along the initial hot isothermal path [tex](Q__H})[/tex] is equal to the work done which is given by the equation;
[tex]Q__H}=nRT__H}In\frac{V_b}{V_a}[/tex]
[tex]Q__H}=nRT__H}In\frac{P_a}{P_b}[/tex]
substituting our data from the given parameters above; we have:
[tex]= 1 * 8.314 * (800+273) *In (\frac{0.2}{60} )[/tex]
= -50882.99 J
To determine the net work produced; we have:
[tex]W_{net}[/tex] = η[tex]Q__H[/tex]
= 0.722 × (-50882.99 J)
= -36737.52 J
∴ The net work produced = -36737.52 J
