Answer:
Explanation:
From the given information:
The pressure of the air during the process = [tex]P_{atm} + P_{due \ to \ wt \ of \ piston}[/tex]
[tex]= 101 \ kPa + \dfrac{75 \ kg \times 9.8 \ m/s^2 \times \dfrac{1 \ N }{1 \ kg.m/s^2} }{\dfrac{\pi}{4}(0.2 \ m)^2} ( \dfrac{1 \ N }{m^2} \times \dfrac{1 \ kPa}{1000 \ n/m^2})[/tex]
The pressure of the air during the process = 124.42 kPa
The boundary work = P × ΔW
The boundary work = 124.42 kPa × (π/4) × (0.2 m)² × 0.25 m × (1 kJ/1 kPa.m³)
The boundary work = 0.977 kJ
The combined work transfer = [tex]W_{boundary} + \Delta U[/tex]
The combined work transfer = 0.977 + 3.109 kJ
The combined work transfer = 4.086 kJ
