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Ideal gas law - work out temperature and pressure

  1. Jan 18, 2010 #1
    A neon tube of volume 1.0 × 10−3m3 holds a sample of 3.0 × 1020 atoms of neon gas in equilibrium. The distribution of speeds of the neon atoms shows a peak at 500 m s−1. The neon may be treated as an ideal gas. (Take the mass of each neon atom to be 20 amu.)

    What is the Temp & Pressure of the gas,
    What is the average kinetic energy of the neon atoms?
    What is the position of the maximum in the energy distribution?


    equations for part 1
    this must be the ideal gas law so
    pV=nRT


    my thoughts so far
    i think i must need to work out how many moles of neon there are but i dont know if thats the correct way to start, there also must be another question as the ideal gas law has two unknowns which means i cant use it. perhaps if i can get Pt 1 then 2 & 3 will be easier ?
     
  2. jcsd
  3. Jan 18, 2010 #2

    Char. Limit

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    There is indeed another equation.

    I think the rms speed equation, which I believe is... u^2=3RT/μ, where u is the rms speed, μ is the molar mass, T is temperature, and R is 8.314.

    Too much information for a homework problem?
     
  4. Jan 19, 2010 #3
    Ok i assume that helps for the second part of the question, but how do i use that before i know the temperature ?

    if anyone can give me a hint i would really appreciate it, i keep going thru my book but i cant find anything to help - im sure its there i just need some advise
     
  5. Jan 19, 2010 #4

    vela

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    Look for a discussion of the Maxwell-Boltzmann distribution in your book.
     
  6. Jan 19, 2010 #5
    I have read that chapter and this is what I think I should be doing?

    E = 1/2mv^2 / 2

    So if I use this equation putting in the figures for mass of the neon atom (20amu) into m and maximum distribution speed 500ms into v1 this will give me E, I guess I would convert the units for mass as amu isn’t an SI unit

    I would then use E = 3/2 kT to find the temperature (obviously rearranging for T)

    I would then use pV=nkT to get the pressure?


    Does this look better?
     
  7. Jan 19, 2010 #6
    The peak of the distribution chart is the most probable speed. Look for an equation which defines most probable speed (Vmp)
     
  8. Jan 21, 2010 #7
    ok thanks for the hint so most probable speed equation is -

    Vmp = SQRT (2kT/m)

    k = boltsman constant 1.38x10^-23

    but i saw this as an alternative form of the equation which was Vmp = SQRT (2RT/M)

    where R = molar gas constant

    how do i know which i use as both are in the book ?
     
  9. Jan 21, 2010 #8

    vela

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    They're the same equation except that one uses the mass m of one particle and the other one uses the mass M of a mole of particles. If you multiply k by Avogadro's number, you get R.
     
  10. Jan 21, 2010 #9
    Ok thanks for that, So i have mass of one atom = 20amu so this is m.

    how do i work out the number of moles or the mass of a mole of particles ?

    sorry for all the Q's but i never really did any chemistry so never learnt any of this before now
     
  11. Jan 21, 2010 #10
    You can google the mass in kg of 1amu, which you can then multiply by 20 to get the mass of one of these atoms. I think the first equation you gave for Vmp involves molecular mass, not molar mass, but it's important to get it right whichever one you use. Make sure you're using the mass of one molecule if that's what the equation requires, or the mass of one mole of the gas if that's what the question requires.
     
  12. Jan 27, 2010 #11
    I'm on a similar question too (the third Q above); How does one find the 'position of the maximum in the energy distribution'?

    Is this the expression I should be using (expression for the energy distribution function)?
    g(E) = C Sqrt E e^−E / kT

    or does the term 'maximum' directly relate to the most probable energy E_mp, in which case should I use E_mp = 1/2kT ?

    The 'maximum' part of the question is, to me at least, a little confusing (although I'm sure it's the key!).

    Any ideas greatly appreciated...
     
  13. Jan 27, 2010 #12
    Hi, I have the same homework as victoriafello, though she's more organised as the deadline is now 10 hours away :-{

    Using the equation for most probable speed

    Vmp = SQRT (2kT/m)

    with very high confidence, from reading round it in the book, that the m involved is molecular rather than molar mass

    the goal would seem to be to rearrange the equation to make T the subject...

    Vmp^2 = 2KT/m

    m*Vmp^2 = 2kT

    (m*Vmp^2) 2k = T

    T = 20amu * (500m s^-1)^2 *(2 *1.38x10^-23)

    T = 5000000 * (2 *1.38x10^-23)

    T = 5.0 *10^-17

    Which is probably a bit low, so hopefully multiplying by the number of neon atoms gives:

    T = (3.0 * 10^20) * (5.0 *10^-17)

    T = 15 * 10^3
    T =



    Ok, so I'm terrible at algebra, still working on this but just thought to put it up in case anyone could let me know if i'm going in the right direction...

    Any help gratefully recieved :)
     
    Last edited: Jan 27, 2010
  14. Jan 27, 2010 #13
    Ok, so clearly simply using amus as units of mass, above, was a bit lame. That's the first thing I'll correct...
     
  15. Jan 27, 2010 #14

    vela

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    First, check your algebra. You solved for T incorrectly. Second, check your units.
     
  16. Jan 27, 2010 #15
    Ok, will do, thanks v much.
     
  17. Jan 27, 2010 #16
    The algebra, incorrect as you pointed out, should be instead:

    Vmp = SQRT (2KT/m)

    Vmp^2 = (2KT/m)

    (Vmp^2) * m = 2KT

    ((Vmp ^2) * m) / 2K = T

    Hopefully.

    Then thinking about the units, the Temperature has to be in Kelvins, as they're the only game in town in this area of physics it seems.

    The speed needs to be in metres per second, which I think is already fulfilled despite it being squared.

    And getting the mass into Kilograms can use the value given in the book for 1 amu = 1.6605 * 10^-27

    I know that there is a physical constant which relates kilograms to kelvins, so having them on each side of the equation feels plausible, especially as there is a constant involved here.

    Putting the numbers in:

    (500m s^-1)^2 *(20(1.6605 * 10^-27)) / 2 *(1.381 *10^-23) = T

    T = 300 K roughly, because I can't yet find a software calculator that doesn't do ridiculous things with the EXP button.

    Thanks for your help Vela, does that look ok?
     
  18. Jan 27, 2010 #17

    vela

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    Yup, looks fine.
     
  19. Jan 27, 2010 #18

    vela

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    You're looking for the value of E that where g(E) hits a maximum.
     
  20. Jan 28, 2010 #19
    I realise I'm looking for the maximum value, but I don't know how to find it... Would you mind elaborating?

    Thanks
     
  21. Jan 28, 2010 #20

    vela

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    You're looking for the maximum of that function, so you find where its derivative is equal to 0.
     
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