# Joule Thomson Effect

1. Jun 12, 2012

### akhyansh

Is the equation
JT = (1/Cp)(2a/RT - b)

2. Jun 12, 2012

### Andy Resnick

The expression you provide appears to be dependent on a specific equation of state (you use Van der Waals, if I may hazard a guess),but the Joule-Thompson experiment is much more general that that.

The Joule-Thompson coefficient is defined as ∂T/∂P at constant enthalpy,and represents a process that is neither isothermal, adiabatic, cyclic, or reversible (throttling of gases). The general expression is

$$(\frac{\partial T}{\partial P})_{H} = \frac{-V}{C_{P}}(1-T\alpha_{T})$$

Historically, the Joule-Thompson experiment led to the development of an absolute temperature scale.

3. Jun 19, 2012

### Carcul

Look also here.

Could you explain that a little more? Because I thought the absolute temperature scale was a consequence of Carnot theorems.

4. Jun 19, 2012

### Andy Resnick

They are related- the experiment was carried out to measure "Carnot's function", which is essentially the efficiency of a Carnot cycle engine (Carnot-Clapeyron theorem).

The details are presented in Truesdell's "The Tragicomical History of Thermodynamics (1822-1854)", specifically section 9D. Briefly, the experiments measured the bath temperature, start and end pressures, and "cooling constant" (Joule-Thomson coefficient), and those were used to fit coefficients in Rankine's equation of state for air p = f(V,T), which would allow the use of air for a 'perfect gas thermometer'.

As a consequence, it became possible to define a temperature scale that is independent of the choice of body used as a thermometer, just as the efficiency of a heat engine is independent of the choice of working fluids.

5. Jun 19, 2012

### Carcul

Thank you very much for your explanation.