July 10, 2020, 04:52:50 AM
Forum Rules: Read This Before Posting

Topic: Help finding partial pressures of a gas mixture added to a room from a fart  (Read 280 times)

0 Members and 1 Guest are viewing this topic.

Offline ShadowDefuse

  • Very New Member
  • *
  • Posts: 1
  • Mole Snacks: +0/-0
We touched on this in gen chem but now we’re covering it again in pchem. I’m wondering about 1a and calculating the partial pressures of the gases in the mixture. What trips me up is that it’s not just the mixture, but also the air pressure already in the room. I’m not sure if I should include that in Ptotal or not.

Not sure how to add pictures but I’m gonna link to imgur and hope that works. https://imgur.com/a/3zPAyne that’s a pic of the problem and my attempt at it.

Thanks in advance

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 2898
  • Mole Snacks: +451/-22
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Well to be honest it's a little ambiguous what "the mixture" is in the problem. Either interpretation for solving part A is defensible, so I'd go with the easier solution (the mixture is just the fart, so to speak).

Also I have to gripe about part B and C. For B, the only gas you'll actually smell is the sulfide, regardless of molecular velocity, since neither of the other gasses have an odor. A bigger issue is found in C: Strictly speaking the dependence you have used (Grahams Law) is the rate of effusion, which is not the same as the diffusion that would occur in the case described in the problem. Effusion is movement of gasses through a small pinhole or porous media, which I don't see as being relevant to this problem, which involves diffusion of the gasses along a concentration gradient. Here the gradient is the limiting factor for the mean rate of gas movement, with the constant of proportionality being the diffusion constant. Although diffusion constants in condensed media are quite dependent on molecular weight (actually, size), they are not so for gasses. For instance, the diffusion constants of methane and carbon dioxide through air at 20 C are 0.21 and 0.16 cm^2/s, respectively. The special case of effusion gives such a tidy expression related to molecular weight because when the pin-hole is very small, the passage of a gas molecular through the hole is related to the probability of the gas molecule making it through the hole before it encounters a collision, and because lighter gasses move faster, their probability of moving through the hole before a collision is greater in proportion to its molecular weight (i.e., velocity). Thus molecular weight limits the mass transfer. When the hole gets very large compared to the mean free path, this relationship no longer holds, and probabilistic behavior of the entire ensemble along the concentration gradient determines the net rate of movement - which is much more complicated and can't be boiled down simply to a ratio of molecular weights. Therefore, which gas makes it to you first, let alone the numerical difference in diffusion times/rates, becomes a complicated expression of the concentration gradient, the diffusion constant, and the geometrical parameters of the diffusion space.

(As I recall, Grahams Law is also only accurate when comparing effusion rates of pure gasses. Its accuracy is lower for mixtures, because, basically, more than one gas is moving through the pinhole at a time, and the gasses may also not be at equal initial concentrations, which affects the relative probability that a particular molecule will encounter the pinhole, irrespective of its relative velocity.)

Anyway, more info than you need and none of this is necessarily a fault of your own, but I thought you may like to know. Basically a very poorly made problem, IMO.
« Last Edit: January 17, 2020, 05:37:30 PM by Corribus »
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Sponsored Links