# Will propane liquefy under these conditions?

Consider a 1:1 mixture of propane and nitrogen at $60°\mathrm{F}$ and $75\:\mathrm{psi}$. Will the propane liquefy if pressure is increased to $200\:\mathrm{psi}$? I am experimenting in the oilfield with a shallow heavy oil lease.

• Is this homework? If so, what materials have you been given to help answer the question? – Abel Friedman Oct 8 '14 at 18:52

First of all, you might want to consider using proper SI units.

## If you know the saturation temperature (boiling point) of propane at a given pressure:

At a pressure of p = 75 psi (0.52 MPa), the saturation temperature of pure propane is 37.13 °F (2.85 °C); thus, at a temperature of 60 °F (16 °C), propane would be gaseous.

At a pressure of p = 200 psi (1.38 MPa), the saturation temperature of pure propane is 104.54 °F (40.30 °C); thus, at a temperature of 60 °F (16 °C), propane would be liquid.

However, a 1:1 mixture of propane and nitrogen contains only 50 % propane; i.e., the mass fraction of propane is wpropane = 0.5. The corresponding amount-of-substance fraction (mole fraction) of propane is xpropane = 0.39.

For ideal gas mixtures, the ratio of partial pressures equals the ratio of the number of molecules:
ppropane = xpropane • p

Therefore, the partial pressure of propane is approximately ppropane = 78 psi (0.54 MPa).

At a pressure of p = 78 psi (1.38 MPa), the saturation temperature of pure propane is 39 °F (4 °C); thus, at a temperature of 60 °F (16 °C), propane would be gaseous.

## If you know the vapour pressure of propane at a given temperature:

At a temperature of 60 °F (16 °C), the vapour pressure of propane is 107.71 psi (0.74 MPa). Thus, at a pressure of p = 75 psi (0.52 MPa), pure propane would be gaseous. At a pressure of p = 200 psi (1.38 MPa), it would be liquid.

However, once again, a 1:1 mixture of propane and nitrogen contains only 50 % propane; i.e., the mass fraction of propane is wpropane = 0.5. The corresponding amount-of-substance fraction (mole fraction) of propane is xpropane = 0.39.

For ideal gas mixtures, the ratio of partial pressures equals the ratio of the number of molecules:
ppropane = xpropane • p

Therefore, the partial pressure of propane is approximately ppropane = 78 psi (0.54 MPa). Since this partial pressure is below the vapour pressure of 107.71 psi (0.74 MPa) at a temperature of 60 °F (16 °C), propane would be gaseous.