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Joule Thomson effect (Porous pluck Experiment)
When a gas under constant pressure is forced through an insulated porous plug through a region of lower constant pressure, its temperature changes. And this effect is called joule Thomson effect.
Experimental setup for joule Thomson effect
Explanation of joule Thomson effect
Let’s replace the porous plug experiments or arrangement buy a simple equivalent arrangement of piston.
- It consists of and insulated cylinder containing a porous plug and filtered with two non conducting piston.
- A fixed mass of gas is filled between the piston one and the plug at a pressure P1 and volume V1.
- The Piston 2 is kept in the other site just behind the plug.
- Both the pistons are simultaneously moved in such way that there is constant pressure of the right hand side of the plug.
- When the gas passed through the plug, gas is expanded to a greater volume V2.
External work done on the gas by Piston 1 is,
∫(v1 to 0) P1. dv = -P1V1
External work done by the gas against the Piston 2,
∫(0 to v2) P2. dv = P2V2
Net external work done by the gas
= P2V2 – P1V1
Since there is no heat exchange between the gas and its surrounding. Then external work must come from the internal energy of the gas. If U1 and U2 are internal energies after gas before and after passing through the plug. Then from first law of thermodynamics we have,
dQ = dU+ qdV = 0
dU = U2- U1
PdV= P2V2 – P1V1
dQ= U2- U1 + P2V2 – P1V1
Constant= U2+ P2V2 = (U1 + P1V1)….(1)
Or H1= H2 ….(2)
∴ (H= U+ PV) enthalpy
In joule Thomson experiment the initial and final enthalpies are same and energy remains constant.
1. Below the Boyle’s temprature,
P2V2 ≥ P1V1
Then, U2 < U1
Hence temprature will fall i.e., it shows cooling effect.
2. Above the Boyle’s temprature
P2V2 < P1V1
Then, U2> U1
Hence, temprature increases i.e., it shows heating effect.
3. At Boyle’s temprature
No changes in the temprature, i.e. there will be No joule Thomson effect.
H= U+PV= constant …(3)
∂H= ∂U + P∂V+ V∂P =0
From first law of thermodynamics,
dQ = dU + PdV …(4)
(dQ = TdS., dS is change in entropy)
TdS= ∂U+ PdV …(5)
From equation 1 and 5,
TdS+ VdP= 0 …(6)
We know entropy is the function of temperature and pressure
By the definition of partial differentiation,
dS= (∂S/∂T)at constant pressure dt+ (∂S/∂P)at constant T dp …(7)
T[ (∂S/∂T)at constant pressure dt + (∂S/∂P)at constant temperature dp]+ VdP= 0 …(8)
We have Cp= (dQ/dT) at constant pressure= T(ds/dt)at constant pressure
(∴ dQ= TdS)
From Maxwell’s equations,
(∂S/∂P)at constant temperature= -( ∂V/∂T)at constant pressure
From equation (8),
Cp dT+ [- T(dV/dT) at constant pressure+ V] dP= 0
dT= (T( dV/dT) at constant pressure – V) dP/Cp
(∂T/∂P)at constant H = 1/Cp (T(dV/dT) at constant pressure – v) =0
μH = (dT/dP) at constant H …(10)
(∴ H remains constant)
It is called joule Thomson constant.
So our article is finished and after completely reading this article, one can easily tell what is Joule Thomson effect. And we have also discussed some more topics like what is basic of joule Thomson effect, Experimental setup of joule Thomson effect, explanation of joule Thomson effect, mathematical expression for joule Thomson effect, relationship between H,T and P, relationship between Cp and Cv etc.
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