Detemining the mass of water vapour in air (Ideal gas law)

AI Thread Summary
To calculate the mass of water vapor in one cubic meter of air at 80% relative humidity and 20°C, the ideal gas law can be applied using the formula p[v] = (eM[w]) / (RT). The vapor density (p[v]) should be measured in kilograms per cubic meter (kg/m³), while the water vapor pressure (e) is expressed in Pascals. The molar mass of water (M[w]) and the gas constant (R) are essential for the calculation, with R being 8.314 J/(mol*K). The temperature must be converted to Kelvin (293.15 K) for accurate results. Understanding these units and conversions is crucial for solving the problem effectively.
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1. I am supposed to calculate the mass of water vapour in one cubic meter of air that has a relative humidity of 80% at 20°C.






2. p[v] = (eM[w]) / (RT)

Where:
p[v] = vapour density
e = water vapour pressure
M[w] = molar mass of water
R = gas constant (8,314 J / (mol*K)
T = temperature (293,15°K)


3. I am struggling with in what unit the vapour density should be measured, as well as the pressure, Pascal? I know this problem is embarrassingly easy, but I've been trying to get the answer for several days, reading up on it online... I miss my old physics-book!

Help would be sincerely appreciated!
 
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Probably it's best if you use SI units. You can take ρ = \frac{m}{V} in \frac{kg}{m^3} and pressure in Pascals.
 

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