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Grade: 12
        
Can anyone please explain Joule-Thomson effect?
I would also like to know about inversion temperature and Joule-Thomson coefficient.
8 months ago

Answers : (1)

Arun
13827 Points
							

The joule-Thomson effect is the change in fluid’s temperature as it flows from a higher pressure region to a lower pressure.

According to thermodynamic principle, the Joule-kelvin effect can be explained best by considering a separate gas packet placed in the opposite flow of direction for restriction. For the gas packet to pass through, the upstream gas needs to perform some work to push through the packet. The work equals to the volume of the packet multiplied by the times of upstream pressure.

W1​​=Vpacket​1​​​​×P1​​

As the packet goes through the restriction, it has to make some room by displacing a considerable amount of the downstream gas. It includes performing the work which equals to the product of packet volume and downstream pressure.

W2​​=Vpacket​2​​​​×P2​​

Due to different effects of compressibility, the work performed upstream is not equal to the amount of work done downstream for real gases. Since depressuring is viewed as an adiabatic process, it reveals that any gas does not exchange work or heat by its surroundings, any change in internal energy has to follow the first law of thermodynamics.

U2​​−U1​​=W1​​−W2​​

Gas molecules are subjected to repulsive and attractive forces, (Van der Waals forces) as they are in random motion. When the gas pressure is lowered, i.e, the average distance between the molecules increases, the attractive forces become dominant for many gases at ambient temperature which results in an elevation in potential energy.

Most of the real gases need more work downstream at ambient temperature, due to the effects of compressibility.

P1​​×V​1​​P2​​×V2​​

The indicates that the internal energy decreases when the gas passes through the restriction.

It can be generalised that for many real gases, the temperature decreases during a reduction in pressure, but not true for every gas and condition. Depressureing is an isenthalpic process which reveals that enthalpy doesn’t change. The temperature can either decrease or increase for any gas based on how the internal energy changes to maintain the enthalpy constant.

8 months ago
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