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falishaduncanovmtz2
06.12.2019 •
Engineering
Agas within a piston–cylinder assembly undergoes an isothermal process at 400 k during which the change in entropy is 20.3 kj/k. assuming the ideal gas model for the gas and negligible kinetic and potential energy effects, evaluate the work, in kj.
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Ответ:
W= 8120 KJ
Explanation:
Given that
Process is isothermal ,it means that temperature of the gas will remain constant.
T₁=T₂ = 400 K
The change in the entropy given ΔS = 20.3 KJ/K
Lets take heat transfer is Q ,then entropy change can be written as
Now by putting the values
Q= 20.3 x 400 KJ
Q= 8120 KJ
The heat transfer ,Q= 8120 KJ
From first law of thermodynamics
Q = ΔU + W
ΔU =Change in the internal energy ,W=Work
Q=Heat transfer
For ideal gas ΔU = m Cv ΔT]
At constant temperature process ,ΔT= 0
That is why ΔU = 0
Q = ΔU + W
Q = 0+ W
Q=W= 8120 KJ
Work ,W= 8120 KJ
Ответ:
Period?
Explanation:
It might just be regular body pain but it normally happens to me on my period