Given g and T, find volume & mass of atmosphere

[brainfuel]

I'm working on building a world, and I have established the mass and radius of the planet, as well as its distance from its star and mean surface temp.

I have determined g (10.33 m/s^2) at sea level, and wonder if this plus the mean surface temp is enough to get a rough idea of the atmosphere's total mass & volume?

Can I just "assign" "appropriate" values for mean atmospheric density and atm pressure at sea level, or will those be dependent on other factors?

TIA
B

 

[brainfuel]

If it helps - I've given the atmosphere a composition with a molar mass of 0.02913351 kg/mol.

I considered the barometric formula to solve this, but I don't have Pb, Tb, or Lb.

 

[Drakkith]

If it helps - I've given the atmosphere a composition with a molar mass of 0.02913351 kg/mol.

I considered the barometric formula to solve this, but I don't have Pb, Tb, or Lb.

I don't know all the math behind this, if the composition of the atmosphere is similar to earth, I'm wondering if you could simply increase the mass by the same proportion as the gravity increased. What do you think?

 

[brainfuel]

I don't know all the math behind this, if the composition of the atmosphere is similar to earth, I'm wondering if you could simply increase the mass by the same proportion as the gravity increased. What do you think?

I've given a composition fairly similar to Earth's - the primary difference is the Nitrogen-Oxygen ratio (instead of 78:21, I made it 69:27, & threw in a couple percent Ne to balance it out). I think I've got it sorted - so I'll dump the math here & see if it seems legit.

molar mass (M) = 0.02914 kg/mol
atmospheric density = 1.28 kg/m^3
depth of atmosphere = 120000 m (compared to 100000 m for 99.99997% of Earth's atmosphere, per Wikipedia)
standard temp (T) = 290.75 K

So for a column of air over one square meter, V = 120000 m^3

120000 m^3 * 1.28 kg/m^3 = 153600 kg of atmosphere per m^2 surface area

153600 kg / 0.02914 kg/mol = 5.271105e6 mol

P = nRT/V = 5.271105e6 mol * 8.314472 Pa m^3 / K mol * 290.75 K / 120000 m^3 = 106187.85 Pa = 106.188 kPa = 1.048 atm

I'm not 100% sure on the density - I've found Earth's atmospheric density at sea level ranging anywhere from 1.2 - 1.29 kg/m^3. The average density of Earth's atmosphere comes out to about 0.099 kg/m^3. It seems like I should probably use that in the equation, but then I come up with a pressure of about 82 kPa, which seems much too low. Not sure if the atmosphere should be significantly thicker than earth's, or if I've got a flaw somewhere else in the process??

 

[IsometricPion]

The wikipedia article on Atmospheric Models is helpful. However, if you have the density, molecular weight, and temperature (all measured) at sea level, (assuming air is an ideal gas) the equation for the pressure at sea level is $$\frac{\rho{}TR}{M}$$ where $$\rho{}$$ is the density, R is the molar gas constant, T is the temperature, and M is the molecular weight. The value I get is about 106.2 kPa. This forumla is just a rearrangement of the ideal gas law.

The average density of Earth's atmosphere comes out to about 0.099 kg/m^3.

Since the pressure in the atmosphere decreases with altitude, so does its density. It turns out not to be a good approximation to assume it is of constant density for the first 100 Km and zero thereafter.

 

[brainfuel]

The wikipedia article on Atmospheric Models is helpful. However, if you have the density, molecular weight, and temperature (all measured) at sea level, (assuming air is an ideal gas) the equation for the pressure at sea level is $$\frac{\rho{}TR}{M}$$ where $$\rho{}$$ is the density, R is the molar gas constant, T is the temperature, and M is the molecular weight. The value I get is about 106.2 kPa. This forumla is just a rearrangement of the ideal gas law.Since the pressure in the atmosphere decreases with altitude, so does its density. It turns out not to be a good approximation to assume it is of constant density for the first 100 Km and zero thereafter.

Okay, that makes sense, and seems to work out to the same result I'd found before. Now I just have to find out if that's a plausible number for the planet in question, which has radius r = 7080 km (1.11 x Earth) and mass m = 7.766 e24 kg (1.3 x Earth). Who knows, I may end up doing all the math over again...