Seeking information on the density of propylene

This post was last edited by ylb913 on 2016-8-16 at 12:46. The saturated vapor pressure of propylene at 70℃ is 3.095 MPa; therefore, under conditions of 70 and 17 kg/cm2, propylene is in a gaseous state. PV = nRT, so V = nRT/P. n = W/M, so W = nM. ρ = W/V, so ρ = nMP/(nRT) = MP/RT. With M = 42, P = 17 + 1 = 18 atmospheres, R = 0.082, and T = 273 + 70 = 343, we get ρ = MP/RT = 26.88 kg/m3. In fact, there are other methods for performing these calculations as well. 1. The density under standard conditions can be easily calculated by dividing the molecular weight by the molar volume of 22.4; 42/22.4 = 1.875, with the unit being grams per liter, or kilograms per cubic meter. Then, based on the formula PV/T = P0V0/T0: (1) Simply put, as pressure increases, density increases; as temperature increases, density decreases. Therefore: 1.875 * (18/1) * (273/343) = 26.86 kg/m3. (2) Derivation process: To calculate the density of a substance under actual operating conditions based on its density at standard conditions, we use the formula ρ/ρ0 = (MP/RT) / (MP0/RT0) = PT0/P0T. Since M/22.4 = ρ0 (the basic unit for molecular weight is g/mol; 1 mole corresponds to 22.4 liters, and dividing these values gives grams per liter, which can be converted to kg/Nm3), with P0 = 1 and T0 = 273, we obtain ρ/ρ0 = P*T0/T. Therefore, ρ = ρ0 * P * T0 / T. 2. There is also a calculation method that compares this value to that of air; the molecular weight of air is 29, and its density at standard conditions is 1.293. Thus, (1.293/29) * 42 = 1.873 kg/m3.