Joule–Thomson Expansion

When a real gas is allowed to flow adiabatically through a porous plug from a region of high pressure to a region of low pressure, its temperature may change. This phenomenon is known as the Joule–Thomson effect, and the process is called Joule–Thomson expansion or throttling.

Porous Plug Experiment

The apparatus used in the Joule–Thomson porous-plug experiment is shown schematically in Fig.

Consider a gas flowing through a thermally insulated tube containing a porous plug P.

Two weightless and frictionless pistons X and Y are fitted on the two sides of the plug. The entire system is thermally insulated; hence, no heat is exchanged with the surroundings.

Initially, let the gas occupy chamber A, having volume V₁, pressure P₁, and temperature T₁. Chamber B is initially empty.

The piston X is pushed slowly so that the gas is forced through the porous plug at approximately constant pressure P₁. On emerging from the plug, the gas enters chamber B and pushes piston Y against a constant pressure P₂.

When the entire gas has passed from chamber A to chamber B, its final volume is V₂.

Work Done

The work done on the gas by piston X is

\[ W_{\mathrm{in}}=P_1\int_{V_1}^{0}dV \] \[ W_{\mathrm{in}}=-P_1V_1 \]

The work done by the gas in pushing piston Y is

\[ W_{\mathrm{in}}=P_2\int_{0}^{V_2}dV \] \[ W_{\mathrm{out}}=P_2V_2 \]

Therefore, the net work done by the gas is

\[ dW=P_2V_2-P_1V_1 \]

According to the first law of thermodynamics,

\[ dQ=dU+dW \]

Therefore,

\[ 0=(U_2-U_1)+(P_2V_2-P_1V_1), \because dQ=0 \]

Hence,

\[ U_1+P_1V_1=U_2+P_2V_2 \]

Therefore,

\[ \boxed{U+PV=\text{constant}} \]

The thermodynamic quantity

\[ \boxed{H=U+PV} \]

is called the enthalpy of the system.

Thus, the essential principle of the Joule–Thomson process is

\[ \boxed{H=\text{constant}} \]

Joule–Thomson Coefficient

For a Joule–Thomson process,

\[ H=U+PV=\text{constant}. \]

Differentiating,

\[ dH=dU+P\,dV+V\,dP=0. \]

From the first and second laws of thermodynamics,

\[ dU=T\,dS-P\,dV. \]

Therefore,

\[ dH=T\,dS+V\,dP=0. \]

Thus,

\[ \boxed{T\,dS+V\,dP=0} \]

Consider entropy S as a function of temperature T and pressure P:

\[ S=S(T,P). \]

Hence,

\[ dS= \left(\frac{\partial S}{\partial T}\right)_P dT+ \left(\frac{\partial S}{\partial P}\right)_T dP. \]

Substituting this in \(T\,dS+V\,dP=0\), we get

\[ T\left(\frac{\partial S}{\partial T}\right)_P dT+ \left[ T\left(\frac{\partial S}{\partial P}\right)_T+V \right]dP=0. \]

We know that, Specific Heat at Constant Pressure

\[ C_P= T\left(\frac{\partial S}{\partial T}\right)_P. \]

Also, from the Maxwell relation,

\[ \left(\frac{\partial S}{\partial P}\right)_T = -\left(\frac{\partial V}{\partial T}\right)_P. \]

Therefore,

\[ C_P\,dT+ \left[ V-T\left(\frac{\partial V}{\partial T}\right)_P \right]dP=0. \]

Hence,

\[ \boxed{ \left(\frac{\partial T}{\partial P}\right)_H = \frac{1}{C_P} \left[ T\left(\frac{\partial V}{\partial T}\right)_P-V \right] } \]

This quantity is called the Joule–Thomson coefficient, denoted by \(\mu_{JT}\).

\[ \boxed{ \mu_{JT} = \left(\frac{\partial T}{\partial P}\right)_H = \frac{1}{C_P} \left[ T\left(\frac{\partial V}{\partial T}\right)_P-V \right] } \]

Case (1) Joule–Thomson Coefficient for an Ideal Gas

For an ideal gas,

\[ PV=RT. \]

At constant pressure,

\[ V=\frac{RT}{P}. \]

Therefore,

\[ \left(\frac{\partial V}{\partial T}\right)_P = \frac{R}{P}. \]

Since

\[ V=\frac{RT}{P}, \]

we obtain

\[ T\left(\frac{\partial V}{\partial T}\right)_P-V = T\frac{R}{P}-\frac{RT}{P}=0. \]

Hence,

\[ \boxed{\mu_{JT}=0} \]

Thus, an ideal gas shows no Joule–Thomson temperature change during throttling.

This is consistent with the fact that the internal energy and enthalpy of an ideal gas depend only on temperature.

Case (2) Joule–Thomson Coefficient for a van-der Waals Gas

For a real gas, the equation of state can be approximately represented by the van der Waals equation:

\[ \left(P+\frac{a}{V^2}\right)(V-b)=RT. \]

Here, a represents the effect of intermolecular attraction, while b represents the finite volume of the gas molecules.

Rearranging,

\[ P=\frac{RT}{V-b}-\frac{a}{V^2}. \]

At constant pressure,

\[ 0= \frac{R}{V-b}\,dT - \frac{RT}{(V-b)^2}\,dV + \frac{2a}{V^3}\,dV. \]

Therefore,

\[ \left(\frac{\partial V}{\partial T}\right)_P = \frac{R/(V-b)} {RT/(V-b)^2-2a/V^3}. \]

For a dilute gas,

\[ b\ll V. \]

Retaining the leading terms, we obtain the approximate relation

\[ \boxed{ \left(\frac{\partial V}{\partial T}\right)_P \approx \frac{V-b}{T} + \frac{2a}{RTV} } \]

Substituting this result into the expression for the Joule–Thomson coefficient,

\[ \mu_{JT} = \frac{1}{C_P} \left[ T\left(\frac{\partial V}{\partial T}\right)_P-V \right], \]

we obtain

\[ \boxed{ \mu_{JT} \approx \frac{1}{C_P} \left( \frac{2a}{RT}-b \right) } \]

The inversion temperature is the temperature at which the Joule–Thomson coefficient becomes zero.

\[ \mu_{JT}=0. \]

Therefore,

\[ \frac{2a}{RT_i}-b=0. \]

Hence,

\[ \boxed{ T_i=\frac{2a}{Rb} } \]

This is the approximate maximum inversion temperature for a van der Waals gas.

Below the Inversion Temperature (Cooling)

If

\[ T\lt T_i, \]

then

\[ \mu_{JT}>0. \]

Above the Inversion Temperature (Heating)

If

\[ T>T_i, \]

then

\[ \mu_{JT}<0. \]

At the Inversion Temperature

If

\[ T=T_i, \]

then

\[ \mu_{JT}=0. \]

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