
Joule-Thomson coefficient is zero at:
A. critical temperature
B. inversion temperature
C. Absolute temperature
D. Baoyle’s temperature
Answer
486.3k+ views
Hint: 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.
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.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Trending doubts
Which one is a true fish A Jellyfish B Starfish C Dogfish class 11 biology CBSE

State and prove Bernoullis theorem class 11 physics CBSE

1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

In which part of the body the blood is purified oxygenation class 11 biology CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells
