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

In summary, to calculate the mass of water vapour in one cubic meter of air with a relative humidity of 80% at 20°C, the equation p[v] = (eM[w]) / (RT) can be used. The vapour density should be measured in kg/m^3 and the pressure should be measured in Pascals, using SI units.
  • #1
Zewz
1
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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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  • #2
Probably it's best if you use SI units. You can take [itex] ρ = \frac{m}{V} [/itex] in [itex] \frac{kg}{m^3} [/itex] and pressure in Pascals.
 

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

1. How is the mass of water vapour in air determined using the Ideal Gas Law?

The Ideal Gas Law, also known as the General Gas Equation, is used to calculate the mass of water vapour in air by taking into account the pressure, volume, temperature, and number of moles of the gas. The formula is as follows: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the temperature. By rearranging the formula to solve for n, we can determine the number of moles of water vapour. Then, by multiplying the number of moles by the molar mass of water, we can calculate the mass of water vapour in air.

2. What is the relationship between temperature and the mass of water vapour in air?

As stated in the Ideal Gas Law, temperature is a crucial factor in determining the mass of water vapour in air. This is because temperature has a direct impact on the pressure and volume of the gas. As temperature increases, the volume of the gas also increases, causing the pressure to increase. This, in turn, increases the number of moles of water vapour, resulting in a higher mass of water vapour in air.

3. How does the pressure affect the mass of water vapour in air?

The pressure is another important factor in determining the mass of water vapour in air. As the pressure increases, the volume of the gas decreases, causing the temperature to increase. This results in a higher number of moles of water vapour and therefore a higher mass of water vapour in air.

4. Can the Ideal Gas Law be used for any type of gas in air?

Yes, the Ideal Gas Law can be used for any type of gas in air, including water vapour. This is because it is a general equation that takes into account the properties of an ideal gas, which includes water vapour.

5. How accurate is the Ideal Gas Law in determining the mass of water vapour in air?

The Ideal Gas Law is a very accurate method for determining the mass of water vapour in air, as long as the gas behaves like an ideal gas. This means that it follows the assumptions of an ideal gas, such as having negligible volume and attractive forces between particles. In real-world situations, these assumptions may not hold true, leading to slight discrepancies in the calculated mass of water vapour. However, for most practical purposes, the Ideal Gas Law provides a very accurate estimate of the mass of water vapour in air.

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