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Thermodynamics Problem involving an ideal gas Helium? please help

  1. Apr 3, 2013 #1
    Hi guys i am new here hope you are all well, had a problem with a physics question hope you can help.

    Helium gas m=2.0(g) is at an initial temperature T1=100 degreees and initial pressure P1=100000 Pa, and has a volume V1. The gas undergoes isobaric (P=constant) expansion until its volume is doubled V2= 2V1. Molar Specific Heat at constant pressure for Helicum Cp=2.08 J/mol K

    a)Final temperature of the gas?

    b)work done by the gas?

    c)the heat input Q into the gas?

    d)What is the change in the thermal energy of the gas (change in E (th) ) ?

    I attempted part a) but i didnt know how i could find the Heat Energy Q needed to change the tempeartue of the ideal gas at constant pressure and i did not know how i could find the number of moles (n) for helium either i know the atomic mass from the periodic table but how to i work out the mass of helium? because n=mass/molar mass, these factors prevented me from applying the formula Q=n*Cp*(Change in T), Could you be specific when showing me the method i need to apply, as my understanding of physics and maths is weak.

    Thank you, very much appreciated
     
  2. jcsd
  3. Apr 3, 2013 #2
    You wrote "Helium gas m = 2.0 (g) ...". If that is not he mass, then what is that?

    Anyway, you do not need anything of the above to solve a). You need to recall how temperature and volume are related in an isobaric process. If you can't recall, use the ideal gas to relate them.
     
  4. Apr 4, 2013 #3

    ehild

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    Homework Helper
    Gold Member

    You know the atomic mass of helium from the Periodic Table, M=4g. You know that the helium gas consists of single helium atoms. You know the mass of the helium gas in the vessel: m=2g. So what is n?

    ehild
     
  5. Apr 4, 2013 #4
    Here is some equations maybe useful for your problem.
    To find:
    1. the Q, just use the Q=mc∆T
    2. the final temp, use P1.V1/T1=P2.V2/T2
    3. the work done, use Q=∆U+w
    4. the mole(n), use n=mass/Molarmass or PV=nRT

    just apply it :)
     
  6. Apr 4, 2013 #5
    I already try found it that for,
    a) the final temperature is 200 degree (don't know whether it is in K or C)
    b)W=415.7 J (another equation, W=P∆V)
    c)Q=0.416 J
    for d) I'm not really understand about the question.

    is it correct? do u have the answer key to check it or not?
     
  7. Apr 4, 2013 #6
    Sorry guys i made an error, yes the mass is 2.0g and i am able to work out the number of moles
     
  8. Apr 4, 2013 #7
    all of them are wrong, these are the answers:

    a)746K
    b)1550K
    c)3879K
    d2329J

    For part a) i now neeed Q but how can i find it if i don't know the specific heat capacity of helium?
     
    Last edited: Apr 4, 2013
  9. Apr 4, 2013 #8

    DrClaude

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    Staff: Mentor

    Have you read the statement of the problem?

    (which, by the way, is not the correct number)

    And as voko said, you don't even need the heat capacity. You can use the ideal gas law instead.
     
  10. Apr 4, 2013 #9
    Okay thanks, sorryi was mistaking the Cp for "c", am going to work it out now and see if i get the correct answer.
     
  11. Apr 4, 2013 #10
    for part a) The final temperature Tf= Q/n*Cp+T1 gave me the correct answer and you were correct the Cp was 20.8 made a typo error.
     
  12. Apr 4, 2013 #11
    I can't work out the work done for part b)? i tried using re arranging PV=nRT to get the volume so i could put it in W=PV but didnt get the correct answer, how did you get the value for volume? and i dont understand the formula Q=U+w, if w is the work and Q what i had worked out from part a) what is "U" thanks
     
  13. Apr 4, 2013 #12

    DrClaude

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    Staff: Mentor

    It's the right way. Please show your calculation if you want some help.

    ##U## is the total energy of the system. Be careful, because there are different conventions for the sign of ##W##. ##U=Q-W## is simply an expression of the first law of thermodynamics.
     
  14. Apr 4, 2013 #13
    V=nRT, i think i made a mistake with the temperature. 4.997x10^-4(n)*8.31(R)*T/1.1x10^5(P)

    Then i put the answer in W=PV

    (For T i used the 747K aswell as the T1 which was given in the problem but still never got the correct answer)

    How would it be possible to use U=Q-W when i don't know what the total energy of the system is, thanks for your help again
     
  15. Apr 4, 2013 #14
    what is the error in my calculation? and did i use the correct value for T? thank you
     
  16. Apr 4, 2013 #15

    DrClaude

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    Staff: Mentor

    Your problem is with ##n##.

    Sorry, the equation should be ##dU = Q - W##. We are considering the change in energy.

    For a monatomic ideal gas, ##U## includes only kinetic energy (no nuclear, electronic, gravitational, etc., energy), so you can even calculate it exactly using
    $$
    U = \frac{3}{2} n R T
    $$
    (You can also calculate it for ideal gases that are not monatomic, but I don't want to go into the details now.)
     
  17. Apr 4, 2013 #16
    how can "n"be wrong if it gave me the correct value for part a)? if it is how can i get the correct value please
     
  18. Apr 4, 2013 #17

    DrClaude

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    Staff: Mentor

    $$
    n = \frac{m}{M} = \frac{2\ \mathrm{g}}{4\ \mathrm{g\,mol}^{-1}} = 0.5\ \mathrm{mol}
    $$
     
  19. Apr 4, 2013 #18
    I got part a) by using the other ideal gas equation PV/T(initial)=PV/T(final), while for V1 I got it from the other PV=nRT equation. After I got the V1, then I used the W=P∆V to find the W. Btw the ∆U is the internal energy, the sum of all K.E of the molecule, this is used at Q=∆U+W (the first law of thermodynamics).
     
  20. Apr 4, 2013 #19
    Btw DrClaude and hbk69, did you change the temperature into Kelvin? Because I'm not convert the 100 degrees because there is no unit provided.
     
  21. Apr 4, 2013 #20
    I just try it again by convert the 100 degrees (found that it is in C) into Kelvin which is 373 K, convert the 100000 Pa into 100 kPa.
    part a), using the PV/T(initial)=PV/T(final):
    100(V1)/373 = 100(2V1)/Tf
    (100 and V1 cancels out)
    Tf=373(2)
    Tf=746 K

    part b), using PV=nRT
    100V1=(2/4) x 8.314 x 373
    V1=15.5 dm^3

    Then, use the W=P∆V
    W=100x15.5
    W=1550J

    for part c) & d) I'm still trying to find it,
     
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