P. Chem. question

[jimmy_beaner]

yeah, I've been working on this for awhile... but could use a little help getting past the stuck point. This is exactly what the question reads:
A star has methane atmosphere. The atmosphere thickness is only 1 m and its temperature is 200ºC. Estimate the mass of this star expressed in Earth’s mass units (5.97 × 1024 kg) using Newton’s law of gravitation.

Our TA later said the problem was more difficult than intended and gave this as a hint.
An average molecule of methane does not have enough kinetic energy to go higher than 1 m. Also assume that the gravitational g for the star is proportional to the mass of the star. Potential energy is mgh, as talked about in class and you can look up the average kinetic energy of a molecule in your notes. DO NOT worry about the radius of the star.

here are some helpful equations in case you don't want to look them up.

P=F/A=(rho)*g*h
F=m*g
Ek = m*v^2/2, Ep = m*g*h
Fg = G (m1*m2)/(r^2)
surface area = 4 (pi) r^2
volume = 4/3 (pi) r^3
Ek = (3*R*T)/(2*Na)

[jimmy_beaner]

also, I could put some of the stuff I've thought about up I suppose if that might help.
Fg = G (m1m2/r^2), but P = F/A, and the force is supplied only by gravity, so:

P = [G m1m2/r^2)/A], also, the area is the surface area of the star, so:

P = [(G m1m2/r^2)/(4 (pi) r^2)] which simplifies to [(G m1m2)/(4 (pi) r^4)]

potentially more in a bit

[jimmy_beaner]

(rho)gh = F/A, thus, (rho)gh = [(G m1m2/r^2)/(4 (pi) r^2)], and then, you can use g = G m1m2/r^2 to cancel those factors leaving (rho)h= 1/4(pi)r^2, or 1=4(pi)(rho)h*r^2

but if I knew the density of the methane, I could find the r^2, because the average height of a molecule of methane will be 0.5 meter (1/2 of 1 m) because of the normal distribution and that leads to the greatest chance of finding a molecule at 0.5m, and the ones higher than that cancel those lower. And after finding the r^2 value, it would be smoother sailing to the end of the problem, but the TA said to not worry about it.

Other things: U = Ep + Ek, U is the internal energy of the system. So you can aruge:
U= m*v^2/2 + m*g*h, and again, you could substitute for g to get:

U = m*v^2/2 + m*h * (G m1m2/r^2)

but also, F = m*a(cceleration)

and you can argue g = GM/r^2; M = g r^2/G, so this of course sets up the proportionality of M to g
_______________________________________________________________________________________
from Avagadro: <Ekin> = 3RT/2 Na, which (after plugging in numbers) yields <Ek> = 9.8023E-21 J, and I'm assuming of course that that is per molecule.
N = n*V, and therefore, we need <Ek> * N or <Ek>*n*V to get the true Ek of the system. Now, we also need to consider the <Ep> of the system, to get the Ep total, then add the two to get the internal energy of the system, which remains constant. Now, Ep = m*g*h. When we think about the mass of methane, 1 mole comes out to 16.05g roughly, and divide that by Na to get the mass of each molecule to be: 2.67E-23g/molecule, and then we need to multiply by the total number of molecules in the system, N.
Ep = 2.67E-23g/molecule *n*V*g*0.5 (as I stated earlier, this is the average height of a molecule of methane).
This also leads nicely to the inclusion of g into the internal energy, U, equation.

U = 1.333E-23*n*V*g + 9.8023E-21*n*V
_______________________________________________________________________________________
U = 1.333E-23*n*V*(GM/r^2)+9.8023E-21*n*V
U = n*V*(8.895E-34*M/r^2 + 9.8023E-21)

U = (3*n*R/2)*T

thus:

5901 n = n*V*(8.895E-34*M/r^2 + 9.8023E-21)
then n cancels from each side:

5901 = V*(8.895E-34*M/r^2 + 9.8023E-21)
ok, I suppose this is how I ended it:
scratch that, how 'bout P1/T1=P2/T2, and one of the sides is standard pressure and temperature.
P2 then equals 175,515 Pa, and another equation:
(rho) = (16.05*175,515)/(8.3145*473.15)
(rho) = 716.068
substituting into M = 3.97034E36 (rho^0.5) gives us:
M = 1.06244E38 grams.... and now divide by the mass of the earth: 5.9742E24 kg or 5.97E27 grams
1.778E10 earth mass units


but in a side note, that's a DAMN big star. (our sun is only 330,000 or so times earth's mass)

[bluemustang]

wow good luck with that buddy

[Kiljosh]

And that ladies and gentlemen, is why I'm not a Chem major.

Sorry I can't help

[lowflyn]

I understood star, atmosphere, and temperature...does that help any?

[jimmy_beaner]

also, from earlier we have 1 = 4(pi)(rho)(h)r^2, which means we can 'solve' for r^2, and it becomes 1/(4(pi)(rho)(h))=r^2, and of course, because we do have volume aspects in our equations it would be nice to have a value for r.
and since our radius isn't negative, that leads only one value:
r = 0.3989/((rho)^1/2)

putting this all together:
M = 1.58377E36/r, r = 0.3989/(rho^.5), and that leads to:
M = 3.97034E36 (rho^0.5)

anyone know how I can find the density of the hypothetical atmosphere?

[jimmy_beaner]

yeah, and sadly, this is the FIRST problem set he's assigned... and he hasn't mentioned ANYTHING about this type of stuff in class, and the book doesn't talk about it either, so it has to be derived all new... and it blows

[Herbstang]

Good Luck with that one only thing I know is on our planet methane is a by-product of biological activity,and in space the methane produced could be linked to volcanic or hydro-thermal activity on your star. methane also oxidises to form water and carbon dioxide.

[jimmy_beaner]

ok, I suppose that's the end of it, but I don't feel good about the answer... not at all. Oh well, I have an answer

[jimmy_beaner]

alright, here's the last one I have:

and seriously, I don't much understand this one, so I may not get an answer, but anyone know?

[ridenmystang]

Yeah I think the answer is 1. It maybe 2 tho.

[Fishers Stang]

U know in the fire service, we have to fig out what psi to use on a dia. of hose x feet long, but i just stuck to "I will put the wet stuff on the red stuff" sorry bro i cant help ya.

[Kiljosh]

Quote: Originally Posted by ridenmystang Yeah I think the answer is 1. It maybe 2 tho.

I was gonna say 3 but 2 is pretty close.

[Rick0636]

Don't know what sort of math that is, but its way over my head, lol.