Joule-thomson coefficient : it represents change in temperature with respect to change in pressure at constant enthalpy.
J.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
J.T.
.
J.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 J.T. <
0 (Joule-thomson coefficient is with negative sign )
(c) isothermal Joule–Thomson coefficient : It is slope of enthalpy vs. pressure at constant temperature
T = (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 ;
J.T. =
(dT / dP )H
T =
J.T.*
Cp
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