In a reversible process both the system and surrondings can be returned to their initial states. a)-True b)-False
1 answer:
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Answer:
The amount of heat transferred to the air is 340.24 kJ
Explanation:
From P-V diagram,
Initial temperature T1 = 27°C
Initial pressure P1 = 100 kPa
final pressure P3 = P2 = 300 kPa
volume at point 2, V2 = V1 = 0.4 m³
final temperature T2 = T3 = 1200 K
To determine the final pressure V3, use ideal gas equation
PV = mRT
Where R is the specific gas constant = 0.2870 KPa m³ kg K
But,
from initial condition, mass m = PV/RT
m = (P1*V1)/R*T1
T1 = 27+273 = 300K
m = (100*0.4)/(0.2870*300) = 0.4646 kg
Then;
Final volume V3 = mRT3/P3
V3 = (0.4646*0.2870*1200)/300
V3 = 0.5334 m³
Total work done W is determined where there is volume change which from point 2 to 3.
W = P3*(V3-V2)
W = 300*(0.5334-0.4) = 40.02 kJ
To get the internal energy, the heat capacity at room temperature Cv is 0.718 kJ/kg K
∆U = m*Cv*(T2-T1)
∆U = 0.4646*0.718(1200-300)
∆U = 300.22 kJ
The heat transfer Q = W + ∆U
Q = 40.02 + 300.22 = 340.24 kJ
Determine the amount of heat transferred to the air, in kJ, while increasing the temperature to 1200 K is 340.24 kJ
The attached file shows the Pressure - Volume relationship (P -V graph)
