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Of course it depends on the target pressure and temperature, so I don't have an exact answer, but I can show you how to make a good guess can form the density of liquid propane and the ideal gas law. The density of the liquid material at average sea-level atmospheric pressure and at its boiling temperature I have looked up and found to be 582 kg/m3. The gas at 25 C and average sea-level atmospheric pressure has a density of (moles*Molecular weight)/Volume = (Pressure*molecular weight)/(ideal gas constant*absolute temperature).

Pressure = 1 atmosphere

Molecular weight = 0.0441 kg/mole

Ideal gas constant = [0.00008206 M3 Atm / K]

Absolute temperature = 298.15 (i.e., 25 C)

or density = 1 Atm * 0.0441 kg/mole / (0.00008206 M3 Atm /K * 298.15 K)

= 1.8025 Kg/M3

Literature seems to say 1.8324 Kg/M3 (which is the same as g/L). This is simply because the ideal gas law is only an approximation -- this gas is close to it liquefaction point, so it is showing some discrepancy from ideal behavior.

The expansion Ratio to get to standard conditions is therefore close to 582/1.80, or ~320

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