
For an ideal gas, Joule-Thomson coefficient is ………..
Answer
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Hint: Joule-Thomson coefficient is used to measure the non-ideality of gases quantitatively. It is a constant which is derived from the Joule-Thomson effect. This process is also known as Kelvin-Joule effect or Joule-Kelvin effect.
Complete step by step answer:
A fluid is something which is either liquids or gases. When a fluid flows from a region of high pressure to a region of lower pressure, there will be a change in temperature. In this process, there does not occur any exchange of heat with the environment.
But this process is only applicable to real fluids. Now let’s define Joule-Thomson coefficient. When the enthalpy, ${{H}}$ is constant, the temperature, ${{T}}$ changes with respect to pressure, ${{P}}$. This rate of change of temperature is known as Joule-Thomson coefficient. It is expressed as ${\mu _{{{J - T}}}}$
This coefficient can also be defined with respect to several other terms like volume, ${{V}}$ of gas, thermal expansion coefficient, $\alpha $ and heat capacity at constant pressure, ${{{C}}_{{p}}}$.
Joule-Thomson coefficient is expressed in the formula given below:
${\mu _{{{J - T}}}} = {\left( {\dfrac{{\partial {{T}}}}{{\partial {{P}}}}} \right)_{{H}}} = \left( {\dfrac{{{V}}}{{{{{C}}_{{p}}}}}} \right)\left( {\alpha {{T}} - 1} \right)$
${\mu _{{{J - T}}}}$ depends on several factors like gas nature, temperature and pressure of gas before expansion.
Now let’s consider ideal gases. When they are expanded, they do not get heated or cooled at constant enthalpy. This denotes that the change in temperature is zero, i.e. $\partial {{T = 0}}$. From the equation given above, it is clear that when $\partial {{T = 0}}$, ${\mu _{{{J - T}}}} = 0$.
Thus the Joule-Thomson coefficient of ideal gases is zero.
Note: Since ideal gases are said to be imaginary, there are no intermolecular forces of attraction between the gas molecules. Thus the internal energy will be zero. The Joule-Thomson effect has both a cooling effect and heating effect. The cooling effect is used in the refrigeration.
Complete step by step answer:
A fluid is something which is either liquids or gases. When a fluid flows from a region of high pressure to a region of lower pressure, there will be a change in temperature. In this process, there does not occur any exchange of heat with the environment.
But this process is only applicable to real fluids. Now let’s define Joule-Thomson coefficient. When the enthalpy, ${{H}}$ is constant, the temperature, ${{T}}$ changes with respect to pressure, ${{P}}$. This rate of change of temperature is known as Joule-Thomson coefficient. It is expressed as ${\mu _{{{J - T}}}}$
This coefficient can also be defined with respect to several other terms like volume, ${{V}}$ of gas, thermal expansion coefficient, $\alpha $ and heat capacity at constant pressure, ${{{C}}_{{p}}}$.
Joule-Thomson coefficient is expressed in the formula given below:
${\mu _{{{J - T}}}} = {\left( {\dfrac{{\partial {{T}}}}{{\partial {{P}}}}} \right)_{{H}}} = \left( {\dfrac{{{V}}}{{{{{C}}_{{p}}}}}} \right)\left( {\alpha {{T}} - 1} \right)$
${\mu _{{{J - T}}}}$ depends on several factors like gas nature, temperature and pressure of gas before expansion.
Now let’s consider ideal gases. When they are expanded, they do not get heated or cooled at constant enthalpy. This denotes that the change in temperature is zero, i.e. $\partial {{T = 0}}$. From the equation given above, it is clear that when $\partial {{T = 0}}$, ${\mu _{{{J - T}}}} = 0$.
Thus the Joule-Thomson coefficient of ideal gases is zero.
Note: Since ideal gases are said to be imaginary, there are no intermolecular forces of attraction between the gas molecules. Thus the internal energy will be zero. The Joule-Thomson effect has both a cooling effect and heating effect. The cooling effect is used in the refrigeration.
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