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Ideal Gas Law: What is the new pressure based on factors?

  1. Oct 22, 2015 #1
    My daughter has the following homework problem we need assistance in confirming if we've understood and completed correctly.

    1. The problem statement, all variables and given/known data

    The Volume goes from 2.00 Liters to 7.00 Liters, the temperature changes from 800 kelvin to 150 kelvin, and the number moles of gas is quadrupled on a gas sample initially at 1.20 atm. What is the new pressure?

    2. Relevant equations

    PV = nRT, where P = pressure, V = volume, n = moles, R = a given number at 0.0821, and T = temperature.

    3. The attempt at a solution

    We assumed that since moles was used here, we need to use the "PV = nRT" version of the ideal gas equation versus the "P1V1 / T1 = P2V2 / T2" version. Given that, we also assumed we'd have to apply the factor of the changes in each applicable variable.

    Volume factor = 7 / 2 = 3.5
    Temperature factor = 150 / 800 = 0.1875
    Moles factor = 4

    Therefore, mathematically, logic stands you can apply the product of those factors to the given starting pressure amount of 1.20 atm to get the new pressure amount.

    PV = nRT
    P = nRT / V
    P = [ (4 n) x (0.0821) x (0.1875 T) ] / (3.5 V)
    P = [ (4 x 0.0821 x 0.1875) / 3.5 ] x (nT / V)
    P = 0.0176 nT / V

    Your factor is therefore 0.0176 which can be applied to the original pressure to get your new pressure.

    Answer:

    P = 1.2 atm x 0.0176
    P = 0.0211 atm

    Can you please confirm if our logic and answer is correct and if not where we went wrong?

    Thanks!!
     
  2. jcsd
  3. Oct 22, 2015 #2
    You can actually use the following:
    (P1xV1)/(n1xT1)=((P2xV2)/(n2xT2)
    it accounts for the change in moles also.

    If you want me to show you why this is, i can show you.
     
  4. Oct 22, 2015 #3
    That's interesting. Didn't know you could do that. Yes, be interested in seeing how you got to there.

    However, I don't think I can use it here because I don't know what the original moles are. I just know they quadrupled.
     
  5. Oct 22, 2015 #4
    Oh ok. Yes, then in that case, since you know the initial temperature, pressure, and volume. You can then solve for the amount of moles. Once you have the answer for the amount of initial moles, you can multiply that by 4, then solve for the final pressure. That's how i would go about it.
     
  6. Oct 22, 2015 #5
    Would you mind demonstrating?
     
  7. Oct 22, 2015 #6
    I solved it in this image:
    Hopefully it's what you're looking for!
     

    Attached Files:

  8. Oct 22, 2015 #7

    Borek

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    Staff: Mentor

    While technically that could be possible, it didn't yield the correct answer (but I have not checked why), and it is the most complicated way of dealing with the problem I can think of.

    Why not split it into two simple problems:

    1. Calculate number of moles present in the initial conditions.

    2. Multiply by 4.

    3. Calculate new pressure knowing n, R, T and V.
     
  9. Oct 22, 2015 #8
    Ah. Makes total sense. Don't know why I didn't think of that approach. Should have kept it simple.

    Still... not sure why mathematically our approach doesn't work since logically you should be able to apply the product of the ratios/factors.

    For example, here's another problem where we used basically the same principals. And we THINK we got it right.

    Question:
    The pressure doubles, the absolute temperate triples, and the number of moles becomes 1/6 as great on a 1.2 L nitrogen sample. What is the new gas volume?

    Our Answer:
    PV = nRT
    V = nRT / P
    V = [ (1/6 n) x (0.0821) x (3 T) ] / (2 P)
    V = [ (1/6 x 0.0821 x 3) / 2 ] x (nT / P)
    V = 0.02025 nT / P
    V = 1.2 L x 0.02025
    V = 0.02463 L

    In this case, you aren't given any values except factors. So, not sure how we could use your approach.

    If I can infringe on a little more of your time... how would you go about that one?

    Thanks
     
  10. Oct 22, 2015 #9
    Yes. This seems like same suggestion as Paul. Thanks to you too!

    Yet.... still not sure how you would apply it to a similar question where we used that answer. See my other reply to Paul.
     
  11. Oct 22, 2015 #10
    Wait! I think I figured it out!

    I don't think I was supposed to include the given amount for "R" at 0.0821 in my approach to figure out the factor amount.

    PV = nRT
    P = nRT / V
    P = [ (4 n) x (R) x (0.1875 T) ] / (3.5 V) <----- Used that instead of: [ (4 n) x (0.0821) x (0.1875 T) ] / (3.5 V)
    P = [ (4 x 0.1875) / 3.5 ] x (nRT / V)
    P = 0.2143 nRT / V
    P = 1.2 atm x 0.2143
    P = 0.2571 atm

    The same answer that Paul got

    This makes sense because

    P = nRT / V

    So the factor would need to be applied to "nRT / V" and not "nT / V" as I did in my original.
     
  12. Oct 22, 2015 #11
    i hope it's not too messy. i used the equation i suggested before. then i just ended up using the coefficients and canceled out the P, n, and T. hope this makes sense.
     

    Attached Files:

  13. Oct 22, 2015 #12

    Borek

    User Avatar

    Staff: Mentor

    0.257 atm is what I got too.

    Sorry, no time for more ATM.

    Edit: and I can confirm 0.3 L for the second problem.
     
  14. Oct 22, 2015 #13
    here's how my equation came up to be. this is how i learned it.
     

    Attached Files:

  15. Oct 22, 2015 #14
    Seems like you got it! Glad you are able to understand it! I'll say goodnight to you and hope to be able to help you in the future!
     
  16. Oct 22, 2015 #15

    Borek

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    Staff: Mentor

  17. Oct 22, 2015 #16
    Awesome. Thanks to both.

    My adjusted approach (which now I know and admit is likely overly complicated) also gets 0.3 L.

    PV = nRT
    V = nRT / P
    V = [ (1/6 n) x (R) x (3 T) ] / (2 P)
    V = [ (1/6 x 3) / 2 ] x (nRT / P)
    V = 0.25 nRT / P
    V = 1.2 L x 0.25
    V = 0.3 L

    I think we get it now.

    Thanks again!

    Going to bed now. =)
     
  18. Oct 22, 2015 #17

    epenguin

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    Homework Helper
    Gold Member

    Put to your daughter the way I would have done it, because I don't like to see young students struggling hard with easy things that they will then forget because they seem complicated. (And OK you do have to know some semi-elementary things for this, dependence on mass density, Boyle's Law, Charles' Laws, meaning of absolute temperature. And OK,Borek has said more or less the same thing,)

    I'd just say: the pressure was so much. Then into the same volume I have pushed 4× as much stuff, mass so pressure is now



    I have then let it expand to 7/2 × what it was, in other words made the density 2/7 × what it was

    2/7 ×

    then I have cooled it to 150/800 of what it was so pressure becomes

    150/800 ×

    what is was after the last operations: in all its pressure has become

    4× 2/7 × 150/800 × than the 1.2 atm it originally was I.e. 0.257 atm , in agreement with others.

    Simple proportions come into so many quantitative chemistry questions. Hope this helps your daughter.

    Not to disdain the equations in the end, have to know about R etc., but helps make plain and less intimidating what they mean.
     
    Last edited: Oct 22, 2015
  19. Oct 22, 2015 #18
    Yes thank you!I knew it was possible to write equations but i didn't know how because it looked highly complicated when ifirst tried! I'll be able to practice and make sure i know what I'm doing now!
     
  20. Oct 23, 2015 #19
    Here is how I derived the equation to solve for final pressure. This show's you where your idea of multiplying by ratios could work if done properly. You dont need to calculate the moles in the beginning, then multiply by 4, then stick new number in.......the moles cancel out as you'll see.....
    plug ur numbers in and see if this gives you the right solution; and remember to watch your units gas problem.jpg
     
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