 # Joule thomson effect notes available in easiest way

Introduction

• In today’s article, we are going to study about ‘ Joule Thomson effect ‘ How beneficial it is and what is the basics of concept of joule Thomson effect you should know before getting a detailed information on it. And hence We will also read about 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 given in Detail.  So, take your notebooks in you hand and get ready to study physics in an easy and sorted way.

# Joule Thomson effect (Porous pluck Experiment)

Definition

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.

# Explanation of joule Thomson effect

Let’s replace the porous plug experiments or arrangement buy a simple equivalent arrangement of piston.

1. It consists of and insulated cylinder containing a porous plug and filtered with two non conducting piston.
2. A fixed mass of gas is filled between the piston one and the plug at a pressure P1 and volume V1.
3. The Piston 2 is kept in the other site just behind the plug.
4. Both the pistons are simultaneously moved in such way that there is constant pressure of the right hand side of the plug.
5. 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

P2V2= P1V1

U2= U1

No changes in the temprature, i.e. there will be No joule Thomson effect.

We have,

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

S= S(T,P)

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.

### Conclusion

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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