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TimeRip496
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I think I can see it mathematically but it is not very intuitive to me as it seems like there is creation of energy?
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For instance,
1) Adiabatic expansion
1st law of Thermodynamics to describe conservation of energy
(-dU) = dQ - W
where dU-change in internal energy, dQ-energy in/out the system, W-work done against the environment
(-dU) as internal energy of system decreases(temperature decreases) due to expansion
Since is adiabatic, dQ = 0, thus dU = W.
2nd law of Thermodynamics to describe direction of natural process
(-dU) = TdS - PdV1
TdS = (-dU) + PdV1
when the system is irreversible, dS > 0. For dS > 0, -dU + PdV1 > 0, thus PdV1 > dU.
Hence PdV1 > W and thus, TdS > Q.
Therefore, increase in entropy for adiabatic expansion.
2) Subsequently, when we compress it adiabatically
Again, 1st law of Thermodynamics to describe conservation of energy
dU = dQ - (-W)
-W as work on system due to compression
Since is adiabatic, dQ = 0, thus dU = W.
2nd law of Thermodynamics to describe direction of natural process
dU = TdS - (-PdV2)
TdS = dU - PdV2
when the system is irreversible, dS > 0. For dS > 0, dU - PdV2 > 0, thus PdV2 < dU.
Hence PdV2 < W and thus, TdS > Q.
Since PdV2 < W and pressure change is the same, dV2 < dV1.
Therefore, we cannot get back to the previous volume at the same temperature.
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Adiabatic means that there is no energy/matter transfer between the boundaries right? The only thing that is allowed is work against environment through expansion or compression.
From my above, why won't the system be such that PdV = W especially since it is adiabatic? Entropy cannot decrease but it can remain the same which means the process can be reversible. The only reason I can think why PdV > W is that the system is not perfect as maybe there are friction? However won't that not be adiabatic? Where did that excess PdV - W comes from?
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