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2. (9 marks) Consider the Joules-Thomson coefficient: What is the Joules-Thomson coefficient for an ideal gas? Explain. What

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Joule-thomson coefficient :  it represents change in temperature with respect to change in pressure at constant enthalpy.

\muJ.T. = (dT / dP )H

(a)  Joule-thomson coefficient is Zero for ideal gases , since it is isoenthalpic process and enthalpy of ideal gases is temperature dependent, so to remain enthalpy constant , temperature has to remain constant.

We have , H = U + PV

To cool a gas, it is allowed to expand without allowing any energy to enter from outside as heat. As the gas expands, the molecules move apart , they do it against the attraction of their neighbours : some kinetic energy is  converted into potential energy for greater separations.
This sequence explains the Joule–Thomson effect: cooling of a real gas by adiabatic expansion. Since ideal gas molecules do not have any attraction , so kinetic energy is lost to overcome attraction for expansion : no energy is lost , thus no cooling and zero is \muJ.T. .

\muJ.T. (ideal gas) = 0

(b) For molecules are condition of net repulsive forces , when repulsive forces are dominant than :    they have a larger molar volume than a perfect gas.

or Z > 1 ( Z = pVm / RT ) (Z = compressibility factor)

So, in Joule–Thomson experiment expansion of gases results in warming , as gas molecules relieved from repulsive forces. Since,  the Joule–Thomson effect results in warming of gas:

or   \muJ.T.  < 0 (Joule-thomson coefficient is with negative sign )

(c) isothermal Joule–Thomson coefficient : It is slope of enthalpy vs. pressure at constant temperature

\muT = (dH / dP )T

We have, H = f (P,T)

dH = (dH / dP )T dP +   (dH / dT )P dT

Joule thomson experiment isoenthalpic, dH = 0

0 = (dH / dP )T dP +   (dH / dT )P ​​​​​​​dT

(dH / dT )P ​​​​​​​dT = -   (dH / dP )T dP

or (dH / dP )T   = - (dT / dP )H (dH / dT )P

we have,   Cp (Heat capacity ) =  (dH / dT )P   ; \muJ.T. = (dT / dP )H

  \muT   = \muJ.T.* Cp

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