Explore isenthalpic expansion with a versatile coefficient tool. Choose models, units, and reporting formats fast. See heating or cooling trends across pressure drops instantly.
Pick a model, enter your state data, and compute μJT with unit-safe outputs.
The Joule–Thomson coefficient is defined for an isenthalpic change as μJT = (∂T/∂P)H.
Example values are illustrative and may not match any specific gas exactly.
| Model | T (K) | Cp (J/mol·K) | Pi → Pf (bar) | Key inputs | μJT (K/bar) | ΔT estimate (K) |
|---|---|---|---|---|---|---|
| Thermodynamic | 300 | 29.1 | 50 → 10 | V=0.60 L/mol, (∂V/∂T)P=0.0020 L/mol·K | 0.82 | -32.8 |
| van der Waals | 300 | 29.1 | 50 → 10 | a=3.60 bar·L²/mol², b=0.0427 L/mol | 0.35 | -14.0 |
| van der Waals | 650 | 29.1 | 50 → 10 | a=3.60 bar·L²/mol², b=0.0427 L/mol | -0.08 | +3.2 |
The Joule–Thomson coefficient, μJT, links temperature change to pressure change during an isenthalpic process. In throttling valves, porous plugs, and capillary restrictions, enthalpy stays approximately constant while pressure drops sharply. The sign of μJT determines whether the gas cools or warms during this expansion.
By definition μJT = (∂T/∂P)H. If μJT is positive, decreasing pressure produces a decrease in temperature. If μJT is negative, the same pressure drop increases temperature. Many common gases at room temperature show positive μJT, while light gases at high temperature can show negative values.
For an ideal gas, internal energy depends only on temperature and enthalpy depends only on temperature. During isenthalpic flow, temperature therefore remains nearly constant, so μJT approaches zero. Real-gas behavior introduces intermolecular forces and finite molecular size, creating nonzero μJT.
The most general input route uses measured or modeled molar volume V and its temperature derivative at constant pressure. The calculator applies μJT = (T·(∂V/∂T)P − V)/Cp. This form is convenient when you have an equation of state or tabulated property data for the gas.
When only simple parameters are available, the low-pressure van der Waals approximation provides a quick estimate: μJT ≈ (2a/(R·T) − b)/Cp. Here a represents attractive forces and b represents excluded volume. The approximation is most useful at modest pressures where higher-order density effects are smaller.
A practical question is whether a throttling step cools the gas. The van der Waals route estimates an inversion temperature Tinv ≈ 2a/(R·b). Below Tinv, μJT tends to be positive and cooling is more likely. Above Tinv, μJT can become negative and heating can occur.
For small to moderate changes, ΔT can be approximated as μJT·ΔP using consistent units. This calculator converts pressure to pascals internally, computes μJT in K/Pa, and then reports μJT in your chosen unit such as K/bar. The trend line summarizes the expected direction for your ΔP.
Use Cp values that match your gas composition and temperature range, because μJT scales inversely with Cp. When using the thermodynamic route, obtain (∂V/∂T)P from the same equation of state used for V. Consistent datasets reduce systematic error and improve comparability across runs.
A positive μJT means temperature decreases when pressure decreases under isenthalpic conditions. For a throttling valve, a pressure drop tends to cool the gas if μJT stays positive over the operating range.
If μJT is negative at the operating temperature, a pressure drop causes temperature to rise. This commonly occurs above the gas’s inversion temperature, where repulsive effects dominate the net real-gas response.
Use the thermodynamic model when you have V and (∂V/∂T)P from an equation of state or tables. Use the van der Waals option for quick, low-pressure estimates when only a and b are available.
Cp appears in the denominator, so larger Cp produces smaller μJT magnitude. Cp also varies with temperature and composition, so matching Cp to your conditions is important for realistic predictions.
It is a useful approximation for small to moderate pressure changes when μJT does not vary strongly. For large drops or near phase boundaries, μJT can change significantly, so detailed property data gives better accuracy.
Use caution. Near saturation or two-phase regions, properties change rapidly and simple approximations can fail. Prefer the thermodynamic route with high-quality property data and interpret results as indicative, not definitive.
Choose the pressure unit that matches your data source. The calculator converts internally to pascals and then reports μJT in your selected unit, so consistent input units for Pi and Pf are what matter most.
Important Note: All the Calculators listed in this site are for educational purpose only and we do not guarentee the accuracy of results. Please do consult with other sources as well.