gchem gas problem

creative8401

Im Anush Hayastan
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Jan 22, 2008
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  1. Pre-Dental
    Hi everybody, i'm having a little trouble with this stp and gas volume

    ok heres what we should be aware with:

    as Temp increases, volume increases, as pressure increases vol decreases

    from kaplan book, they give an example of what seems to contradict this fact:

    What is the molar mass of a 2L sample of gas that weighs 8g at a temp of 15C and a pressure of 1.5atm?

    They do: V = (2L)(273/288)(1.5/1)

    What they are doing is comparing the volume of the gas with the volume at STP (cool I get that), but why are they putting an increase in temp in the denominator? (as temp increases, volume increases is a fact) same for pressure (its reversed!) any insight would be great!

    thnx!!
     

    eatkabab

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    Apr 20, 2008
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      did you not write some stuff? I don't completely understand what kind of a problem it is. it seems like a gas density problem which you'd use the equtn P(mm)=DRT, but you've written an eqtn that is different.

      would this happen to be correct?

      1.5X=(8/2)(.1)288 X=mm=about 77g/mol ?


      edit:
      use real R instead of rounded R (.08206 instead of .1)

      P(mm)=DRT
      1.5(mm)=(8/2)(.08206)288 (mm) = 63g/mol
       

      eatkabab

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      Apr 20, 2008
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        oh yeah, that makes sense now.

        they are converting the gas to STP so that they can compare it to an ideal gas at STP which would take up 22.4L/mol.

        so the first thing they do is PV/T = PV/T but the final P is just 1 so they didn't write it. once they've got the volume at STP, they find the density at STP which would be grams/vol = 8/2.84 = 2.82 g/L.

        as usual, you treat it as an ideal gas, so you can compare it to one. ideal gases take up 22.4L/mol at STP. and since we have our gas at STP, we can just do the dimensional analysis to get the g/mol: 2.82g/L (22.4L/mol) = 63.2g/mol.


        but this is a very roundabout way to get the answer. you don't need to convert everything to STP in order to get answers. that would just suck for lab people...

        my original way works too. the reason I didn't get the exact answer as in the book is cuz I used .1 for R instead of .08206. here's the same method I used at the beginning with the unrounded R:

        P(mm)=DRT
        1.5(mm)=(8/2)(.08206)288 (mm) = 63g/mol
         
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        dentalplan

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        Mar 1, 2008
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        1. Pre-Dental
          Here is exactly what they have:

          What is the molar mass of a 2L sample of gas that weighs 8g at a temp of 15C and a pressure of 1.5atm?

          Density = 8g/2L at 15C and 1.5atm

          V = (2L)(273/288)(1.5/1) = 2.84L

          8g/2.84L = 2.82 g/L at STP

          (2.82g/L)(22.4L/mol) = 63.2g/mol


          The way they explained it is COMPLETELY unneccesary. THE hardest way possible to explain this problem.

          The easiest way to do this problem is to simply apply the ideal gas law: PV=nRT

          (1.5) (2) = :thumbdown: (.0821) (288)

          solve for n. and n= 0.126878

          there are 8 grams. 8 divided by 0.126878 equals 63.2 because g/mol is the units of molar mass.

          Another way to do this problem is to use this equation: density= (MM)(P)/RT.....which is actually derived from the ideal gas law. Either way, it's the same thing, and the same concept.

          Does that clear anything up?-or were you thinking of something more specific?
           

          dentalplan

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          Mar 1, 2008
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          1. Pre-Dental
            Hi everybody, i'm having a little trouble with this stp and gas volume

            ok heres what we should be aware with:

            as Temp increases, volume increases, as pressure increases vol decreases

            from kaplan book, they give an example of what seems to contradict this fact:

            What is the molar mass of a 2L sample of gas that weighs 8g at a temp of 15C and a pressure of 1.5atm?

            They do: V = (2L)(273/288)(1.5/1)

            What they are doing is comparing the volume of the gas with the volume at STP (cool I get that), but why are they putting an increase in temp in the denominator? (as temp increases, volume increases is a fact) same for pressure (its reversed!) any insight would be great!

            thnx!!


            What they did here is find the volume of the gas at STP and the conditions the original gas was in.

            You can find the volume of the gas by doing an initial/final state problem:

            PV=nRT (STP)
            --------
            PV=nRT (original)


            (1)(V)=nR(273)
            --------------
            (1.5)(2)=nR(288)

            n and R cancel out.

            It becomes V/3 = 273/288

            Solve for V.

            V (at STP)= 2.84375 L

            That's how they got that V. Now if you keep going..8g/ 2.84375 L =
            2.813 g/L: the density of the gas at STP.

            Plus since we know that there are 22.4 mol/L of any gas at STP, multiply 22.4 by 2.813 so you can get rid of the L values. That step is key to understanding. You multiply them to get rid of L, and your answer is in terms of g/mol (the units of molar mass). So, 22.4 x 2.813 = 63.015 g/mol

            So yea, there are a few ways to do this problem. Absolutely no reason to do more work than you have to do though...(meaning you don't even need to worry about STP at all)....just apply the ideal gas law. That way is faster and easier to comprehend.
             
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