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The ideal gas law states that: DENSITY = P / TR (pressure(P) divided by temperature(T) multiplied by a constant(R)) From the average values found in our atmosphere, we can use a temperature at about 2 km altitude of 275.15K (+2.15C) and a pressure at 2 km of 79.495KPa (795 mb). For dry air, the constant(R) equals 287 J/K*kg. This gives us a density of 1.007 kg/m┬│ for dry air. For pure water vapor, we would use a constant(R) of 461 J/K*kg which ends up giving us a density for pure water vapor of 0.627 kg/m┬│. The density of the cloud droplets in the cloud comes out to .0005kg/m┬│. Because the cloud is made up of a small amount of water vapor and a large amount of air, we would need to calculate the partial pressure of the water vapor, which in this case comes out to around 7 mb, or 0.9% of the total pressure (795 mb). So, [99.1 X (dry air density) + 0.9 X (moist air density)]/100 is approximately the density of the cloud itself. Calculating this gives a density in the cloud of 1.003 kg/m3, compared to 1.007 kg/m3 in the dry air surrounding the cloud. |