Deriving an equation for the density of air

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SUMMARY

This discussion focuses on deriving an equation for the density of dry air based on the density of moist air, relative humidity, saturated steam pressure, and total pressure. The ideal gas law (PV = nRT) is utilized, along with the equations for partial pressures and relative humidity. The final derived equation for the density of dry air is ρ_i = (M_iP_i)/(M_i^'RT - P_vV_v + P_iP_v), where M_i is the molar mass of dry air, M_i^' is the molar mass of moist air, P_i is the partial pressure of dry air, and P_v is the partial pressure of water vapor.

PREREQUISITES
  • Understanding of the ideal gas law (PV = nRT)
  • Knowledge of partial pressures in gas mixtures
  • Familiarity with concepts of relative humidity and saturation pressure
  • Basic algebra for rearranging equations
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sickboy
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First of all, I'm new to this forum, so please excuse any newbie etiquette errors I have made =)

My problem is this: I'm doing a lab report on an experiment where we attempted at measuring the density of air. I won't go to the details, but basically we just measured air on different pressures.

The task is to, given the following equations, derive an equation for the density of dry air as a function of the density of moist air, relative humidity, pressure of saturated 'steam' (can't figure a better translation..Oh yeah, I'm a Finn :) ) and the total pressure.

The first equation:
<br /> \frac{M_i^&#039;}{M_i}=\frac{\rho_i^&#039;}{\rho_i}=1-\frac{M_i-M_v}{M_i}*\frac{e}{p}\approx 1-0,378\frac{e}{p}<br />

...and the second:
<br /> f=\frac{F}{F_m}*100\%=\frac{e}{e_m}*100\%<br />

Notations:
M_i = molemass for completely dry air
M_i^&#039; = molemass for moist air (air with some water vapour in it)
rhos are densities of dry & moist air, respectively
M_v = molemass for water
e = partitial pressure of water vapour
p = total pressure
f = relative humidity
e_m = the pressure required for the condensation of water ( again, not so sure of the translation)
F = absolute humidity (measured in, for example, g/m^3)I hope I didn't forget anything. I have been trying to solve this with my mates for days and days now, and I'd really appreciate any help...

Oh yeah, and I'm sorry, but I can't provide you with any of my sketches, as all I have are red-rimmed eyes and a heap of trash paper =)
 
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Hi there, welcome to the forum! Don't worry about making newbie etiquette errors, we're all here to learn and help each other out.

To derive an equation for the density of air, we can start by looking at the ideal gas law, which states that:

PV = nRT

where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature.

In this case, we are dealing with a mixture of dry air and water vapor, so we need to take into account the partial pressures of each component. We can rewrite the ideal gas law as:

P_iV_i + P_vV_v = n_iRT + n_vRT

where P_i and P_v are the partial pressures of dry air and water vapor, respectively, and V_i and V_v are the corresponding volumes.

We can also rewrite the equation for relative humidity as:

f = e/e_m * 100%

where e_m is the pressure required for the condensation of water vapor, which we can assume to be constant for a given temperature.

Now, let's substitute in the values given in the problem:

P_iV_i + P_vV_v = n_iRT + n_vRT

P_iV_i + P_vV_v = M_i^'RT + M_vRT

where M_i^' and M_v are the molar masses of moist air and water vapor, respectively.

Next, we can rearrange the equation to solve for the volume of dry air:

V_i = (M_i^'RT - P_vV_v)/(P_i - P_v)

To calculate the density of dry air, we divide the mass of dry air by its volume:

\rho_i = M_i/V_i

Substituting in the equation for V_i, we get:

\rho_i = M_i/[(M_i^'RT - P_vV_v)/(P_i - P_v)]

Simplifying and rearranging, we get the final equation for the density of dry air:

\rho_i = (M_iP_i)/(M_i^'RT - P_vV_v + P_iP_v)

I hope this helps! Let me know if you have any questions or if you need me to clarify anything. Good luck with your lab report!
 

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