Question

Joule-Thomson coefficient is zero at:
A. critical temperature
B. inversion temperature
C. Absolute temperature
D. Baoyle’s temperature

Answer 1

we will substitute the value of critical temperature, inversion temperature and Boule temperature in Joule- Thomson coefficient formula and determine the temperature at which it becomes zero. The critical temperature is the temperature, below which a gas can be liquefied. Inversion temperature is the temperature, at which the value of van der waals constant a, and b becomes equal.

Complete step by step solution:
In the joule-Thomson experiment, a gas expands from initial high pressure to final low pressure under adiabatic condition.
The value of joule-Thomson coefficient is as follows:
$\,{\left( {\dfrac{{dP}}{{dT}}} \right)_H} = \,\dfrac{1}{{{C_P}}}\left[ {T{{\left( {\dfrac{{dV}}{{dT}}} \right)}_P} - V} \right]$
$\Rightarrow {\left( {\dfrac{{dP}}{{dT}}} \right)_H}$is known as Joule-Thomson coefficient ${\mu _{JT}}$.
$\Rightarrow {\mu _{JT}}\, = \,{\left( {\dfrac{{dP}}{{dT}}} \right)_H}$
$\Rightarrow {\mu _{JT}}\, = \,\dfrac{1}{{{C_P}}}\left[ {T{{\left( {\dfrac{{dV}}{{dT}}} \right)}_P} - V} \right]$
For the van der waals gas the value of Joule-Thomson coefficient is as follows:
${\mu _{JT}}\, = \,\dfrac{1}{{{C_P}}}\left[ {\dfrac{{2a}}{{RT}} - b} \right]$
The formula of critical temperature is as follows:
${{\text{T}}_{\text{C}}}\,{\text{ = }}\,\dfrac{{{\text{8a}}}}{{{\text{27}}\,{\text{Rb}}}}$
So, when ${\text{T}}\,{\text{ = }}\,{{\text{T}}_{\text{C}}}$
${\mu _{JT}}\, = \,\dfrac{1}{{{C_P}}}\left[ {\dfrac{{34\,aRb}}{{R8a}} - b} \right]$
$\Rightarrow {\mu _{JT}}\, \ne 0$
The formula of inversion temperature is as follows:
Inversion temperature is the temperature, below which if the gas is allowed to expand it shows the cooling effect and above the inversion temperature gas shows the heating effect so, it is not the highest temperature at which vapour pressure of a liquid can be measured.
${{\text{T}}_{\text{i}}}\,{\text{ = }}\,\dfrac{{{\text{2a}}}}{{{\text{Rb}}}}$
So, when ${\text{T}}\,{\text{ = }}\,{{\text{T}}_{\text{i}}}$
${\mu _{JT}}\, = \,\dfrac{1}{{{C_P}}}\left[ {\dfrac{{2aRb}}{{R2a}} - b} \right]$
${\mu _{JT}}\, = 0$
The formula of Boyle’s temperature is as follows:
${{\text{T}}_{\text{C}}}\,{\text{ = }}\,\dfrac{{\text{a}}}{{{\text{Rb}}}}$
So, when ${\text{T}}\,{\text{ = }}\,{{\text{T}}_{\text{C}}}$
${\mu _{JT}}\, = \,\dfrac{1}{{{C_P}}}\left[ {\dfrac{{2aRb}}{{Ra}} - b} \right]$
$\Rightarrow {\mu _{JT}}\, = \,\dfrac{b}{{{C_P}}}$

So, Joule-Thomson coefficient is zero at inversion temperature.

Therefore, option (B) inversion temperature is correct.

Note: Joule-Thomson process is an isenthalpic process. For ideal gas according to the ideal gas equation, the value of ${\left( {\dfrac{{dV}}{{dT}}} \right)_P}$is,
$pV = \,nRT$
${\left( {\dfrac{{dV}}{{dT}}} \right)_P} = \,\dfrac{{nR}}{p}{\left( {\dfrac{{dT}}{{dT}}} \right)_P}$
${\Rightarrow \left( {\dfrac{{dV}}{{dT}}} \right)_P} = \,\dfrac{{nR}}{p}$
On substituting the value of ${\left( {\dfrac{{dV}}{{dT}}} \right)_P}$ in ${\mu _{JT}}\, = \,\dfrac{1}{{{C_P}}}\left[ {T{{\left( {\dfrac{{dV}}{{dT}}} \right)}_P} - V} \right]$,
${\mu _{JT}}\, = \,\dfrac{1}{{{C_P}}}\left[ {T\dfrac{{nR}}{p} - V} \right]$
$\Rightarrow {\mu _{JT}}\, = \,\dfrac{1}{{{C_P}}}\left[ {\dfrac{{pV}}{p} - V} \right]$
$\Rightarrow {\mu _{JT}}\, = \,0$
So, for ideal gas, the Joule-Thomson coefficient is always zero. For real gas it becomes zero at inversion temperature.