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Two views of atmospheric pressure

January 13, 2013

When we cover air pressure we usually develop the ideal gas law based on conceptual arguments and experiments with the PhET Gas Properties simulation. The ideal gas law relates pressure to volume, temperature, and the number of gas particles. In the same chapter, we will also introduce the concept of atmospheric pressure and relate this pressure to the weight of a column of air. At first glance, these two views of air pressure seem rather disjoint. This has bothered me for a while but thankfully once I sat down to work it out, I found that I can relate these two views.

I first tried to calculate the value for atmospheric pressure (about 15 lbs/in2) using the ideal gas law. I know that the mass density of air at ground level is about 1 kg/m3 (air is about 1,000 times less dense than water). I can relate mass density to the number of particles in a given volume according to ρ=Nm/V where m is the mass of a single air particle. Air is mostly N2 and 1 mole of N2 has a mass of 0.028 kg so I will use m=30*10-3/6*1023=5*10-26 kg. Using ρ = 1 kg/m3 and my value for m I find that at ground level N/V is about 2*1025 m-3.

I know Boltzmann’s constant is about 10-23 kg*m2/s2*K and room temperature is around 300 K. Putting this all together I get P=(N/V)kBT=(2*1025 m-3)(10-23 kg*m2/s2*K)(300 K) so P is about 60,000 N/m2. This is about 9 lbs/in2 – close enough to convince me that this calclation works.

OK, so I can use the ideal gas law to get a value for atmospheric pressure. Can I relate this to the weight of the atmosphere? Thinking about the weight of the atmosphere, I can write the pressure at ground level as P=g∫0 ρ(h) dh. The density will fall off exponentially with height and can be written as

ρ(h)=ρ0e-mgh/kBT where ρ0 is the mass density at ground level. Taking the integral I get P=g(kBT/mg)ρ0. Writing ρ0 as Nm/V I get P=NkBT/V, the ideal gas law.

I still don’t have a real nice conceptual argument for why the two views are consistent. I can say that the amount of atmosphere above my head (and hence its weight) depends on the temperature and the density of air particles at ground level (together these determine how many particles have enough kinetic energy to make it to a given height) but I’d like something a little cleaner. At the very least, at least I’ve convinced myself that the two views of atmospheric pressure are mathematically equivalent.

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