what is heat transfer if the internal energy is decreased by 40kj.kg
Answers
Answer:
Given A mass of 8 kg of gas expands within a flexible container so that the p-v relationship is of the form pv1.2 = constant. The initial pressure is 1000 kpa and the initial volume is 1 m3. The final pressure is 5 kpa. If specific internal energy of the gas decreases by 40 kj/kg, find the heat transfer in magnitude and direction.
We can get the equati2 / non as P2/P1 = (V1/V2)^n
V2/V1 = (P1/P2)^1/n
V2 = (P1/P2) ^1/n x V1
here V1 = 1, P1 = 1000, P2 = 5, n = 1.5, substituting this we get
V2 = (1000 / 5)^1/1.5 x 1
V2 = 82.7 m^3
Now W = P1V1 - P2V / n - 1
= 1000 x 1 = 5 x 82.7 / 1.2 - 1
W = 2932.5 K J
ΔE = m x specific internal energy
= 8 x - 40 = - 320 K J
Q = ΔE + W
= - 320 + 2932.5
Q = 2612.5 K J
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