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Finding the pressure of a gas given a density?

  1. Oct 2, 2013 #1
    1. The problem statement, all variables and given/known data
    Ethanethiol (CH3CH2SH) can be smelled in as little as 0.36 parts per billion. 1 ppb = 1x10^-9 g/mL. What is the pressure of the gas in mmHg at this concentration?


    2. Relevant equations
    None


    3. The attempt at a solution
    Well, I took .36 of 1x10^-9 g/mL. Not quite sure what to do next. I know the density could also be expressed as (Pressure)(Molar mass)/(Gas constant)(Temp), but I'm not given a temperature. A nudge in the right direction would be greatly appreciated!
     
  2. jcsd
  3. Oct 2, 2013 #2

    DrClaude

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    Personally, I wouldn't know how to solve it without making a series of additional assumptions, like treating all gases as ideal and setting a value for atmospheric pressure, if a numerical result is to be given. Is what you wrote the exact and complete problem statement?

    In the mean time, the only nudge I can give you is: think partial pressures.
     
  4. Oct 2, 2013 #3
    Well, I omitted some extra stuff that I thought to be irrelevant to the problem, but here it is in full:

    According to the 2000 edition of the Guinness Book of World Records, ethanethiol (CH3CH2SH) is the "smelliest substance" in existence. It can be smelled in as little as 0.36 parts per billion. 1 ppb = 1x10^-9 g/mL. What is the pressure of the gas in mmHg at this concentration?

    Thanks for the partial pressure hint, but it doesn't appear that I'm given a total pressure or anything else to work with. :frown: There must be a way to the answer through that density though, right (That's all I can think of)?
     
  5. Oct 2, 2013 #4

    SteamKing

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    I don't think it would be unreasonable to use standard pressure and temperature in this case. I'm sure the Guiness people wouldn't complain.
     
  6. Oct 2, 2013 #5
    Thanks for the tips! I used 273 K and I got 9.9x10^-5 mmHg. Is this right?

    I used the equation Pressure = (density)(gas constant)(temp)/(molar mass)
     
    Last edited: Oct 2, 2013
  7. Oct 3, 2013 #6

    Borek

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    That's not what I got. One of us is wrong.
     
  8. Oct 3, 2013 #7
    Ah... I've been sitting here trying to figure out what I did wrong but I'm not sure... These are the steps I took though:

    1.) 0.36 of 1x10^-9 g/mL = 3.6x10^-10 g/mL
    2.) Converted this to g/L = 3.6x10^-7 g/L
    3.) Plugged it into the equation from my other post: (3.6x10^-7 g/L)(0.08206 L*atm/mol*K)(273 K)/(molar mass of ethanethiol = 62.15 g/mol) = 1.3x10^-7 atm
    4.) Converted atm to mmHg to get 9.9x10^-5 mmHg
     
  9. Oct 3, 2013 #8

    Borek

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    OK, I somehow missed 273K and started with a different definition of STP, hence the difference.
     
  10. Oct 3, 2013 #9
    Oh, no problem! Yeah, I used the values my textbook gave me for STP. Did we still reach our answers in the same way (or maybe in similar ways)?
     
  11. Oct 4, 2013 #10

    Borek

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    In equivalent ways - I found number of moles of the gas in 22.4 L (concentration*volume -> mass -> moles), then multiplied by 760 (mmHg). 22.4 is a volume of 1 mole of gas at STP, pressure is directly proportional to number of moles.
     
  12. Oct 4, 2013 #11
    Dr. Claude has it right. Sounds like a partial pressure problem. The total pressure is the sum of the partial pressures for each gas in the mixture. In this case, it is air (N2, O2, ...) + ethanethiol. I would assume ideal gas behavior (pV = nRT) and work from there.
     
  13. Oct 4, 2013 #12

    Borek

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    Let me put it differently - Dr. Claude suggestion points to one of many equivalent ways of solving the problem.
     
  14. Oct 4, 2013 #13

    DrClaude

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    Exactly. I didn't see buddhaskillet's point in posting a reply to a solved problem.
     
  15. Oct 4, 2013 #14
    Maybe I'm missing something. It doesn't matter what the temperature is, and, if we are smelling the gas, then it must be in air. I think it was meant for the OP to assume that the total pressure is 1 atm. When concentrations of contaminants in air are given by regulatory agencies and atmospheric scientists, they are given in ppbv, which is the same thing as mole fraction. So the mole fraction of the CH3CH2SH is 0.36 x 10-9, and its partial pressure is 0.36 x 10-9 atm.
     
  16. Oct 4, 2013 #15

    DrClaude

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    I was also bothered by the definition in the OP of 1 ppb = 1x10^-9 g/mL, which doesn't make any sense, but I didn't want to take up the issue for fear of interfering with the poster's homework.
     
  17. Oct 4, 2013 #16
    Hey everyone, thanks for responding. I found everyone's posts to be very informative, and it got me thinking about the different ways in which I could have solved this problem. I'll bring this up in my prof's office hours next week and ask her how she intended for us to solve the problem.

    RE: the ppb definition, that was just a given from my worksheet. I'll ask her about it! Once again, thanks to everyone for providing their input, I appreciate it!
     
  18. Oct 5, 2013 #17

    Borek

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    ppb given as 10-9 g/mL makes sense for water solutions, where you can assume 1 mL to have a mass of around 1g. Most likely this approach was blindly applied here.
     
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