For an ideal gas, the value of the Joule-Thomson coefficient is

The correct answer is zero.

The Joule-Thomson coefficient (${\mu }_{\mathrm{JT}}$) is a thermodynamic parameter defined as the rate of change of temperature ($T$) with respect to pressure ($P$) during an isenthalpic (constant enthalpy, $H$) expansion. Mathematically, it is expressed as:

${\mu }_{\mathrm{JT}}={\left(\frac{\partial T}{\partial P}\right)}_H$

Using thermodynamic relations, the Joule-Thomson coefficient can also be written in terms of the constant-pressure heat capacity ($C_p$), temperature ($T$), volume ($V$), and the coefficient of thermal expansion as follows:

${\mu }_{\mathrm{JT}}=\frac{1}{C_p}\left[T{\left(\frac{\partial V}{\partial T}\right)}_P-V\right]$

For one mole of an ideal gas, the equation of state is given by:

$PV=RT$

where $R$ is the universal gas constant. Rearranging this equation for volume ($V$) gives:

$V=\frac{RT}{P}$

Differentiating $V$ with respect to temperature $T$ at constant pressure $P$ yields:

${\left(\frac{\partial V}{\partial T}\right)}_P=\frac{R}{P}$

Substituting this derivative back into the expression for the Joule-Thomson coefficient:

${\mu }_{\mathrm{JT}}=\frac{1}{C_p}\left[T\left(\frac{R}{P}\right)-V\right]$

Since $\frac{RT}{P}=V$ for an ideal gas, the equation simplifies to:

${\mu }_{\mathrm{JT}}=\frac{1}{C_p}\left[V-V\right]=0$

Conceptually, this is because the molecules of an ideal gas have zero intermolecular forces of attraction or repulsion. When an ideal gas expands, no work is done against intermolecular forces, and therefore, there is no change in internal energy or temperature. Thus, the Joule-Thomson coefficient of an ideal gas is always zero.