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Homework Help: Another homework question

  1. Sep 10, 2010 #1
    I'm really starting to hate this class, the lecture didn't even begin to explain any of this I have no idea where to begin. We've been working on change in internal energy and work done on/by a gas, etc, and we get a question like this on the homework:

    Water standing in the open at 29.7°C evaporates because of the escape of some of the surface molecules. The heat of vaporization (518 cal/g) is approximately equal to εn, where ε is the average energy of the escaping molecules and n is the number of molecules per gram. (a) Find ε, in calories. (b) What is the ratio of ε to the average kinetic energy of H2O molecules, assuming the latter is related to temperature in the same way as it is for gases?


    I don't even know where to begin?
     
  2. jcsd
  3. Sep 10, 2010 #2

    kuruman

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    Begin by calculating the number of water molecules in one gram of water.
     
  4. Sep 10, 2010 #3
    that'd be .05549, right?
     
  5. Sep 10, 2010 #4

    kuruman

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    Are you saying that one gram of water has .05549 molecules in it? If so, then one molecule would have a mass of 18 grams and you would need about 10 water molecules to quench your thirst. Does that sound right?
     
  6. Sep 10, 2010 #5


    i guess i was trying to find the moles? or something, is it not

    n = Msample / M

    where M = molecular mass = 18.02g for H2O?

    i guess i only got halfway, that would be the number of moles, correct?

    then to get the number of molecules you multiply that by Avo's Number

    (1/18.02mol) * (6.022e23/mol) = 3.34e22 that sound closer?
     
  7. Sep 10, 2010 #6

    kuruman

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    Correct. Can you find ε now? You know that 3.34e22 molecules require 518 cal to escape, so how many calories (on the average) does one molecule require to escape?
     
  8. Sep 11, 2010 #7
    I'm honestly not quite sure

    it says En = the heat of vaporization (518cal/g) where n is the molecules per gram. so that means En = 518cal/g, n = 3.3418e22 (whats the unit for this, mols?), then would E = 518cal/g / 3.3418e22 ?

    that doesnt sound right seem's like an incredibly small number... 1.55e-20


    edit: trying something right now ill post up my answer...

    ok so correct me if this is wrong--

    since 1g = 3.3418e22 molecules, then 1 molecule = m / 18.02 * 6.022e23

    1molecule * 18.02 / 6.022e23 = 2.99e-23, meaning 1 molecule is 2.99e-23g?
     
    Last edited: Sep 11, 2010
  9. Sep 11, 2010 #8

    kuruman

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    Indeed it is an incredibly small number. Don't forget, however, that 3.34e22 molecules is an incredibly large number. How many calories do you think you would have to give to one of these molecules if you just had 518 cal to give away?
     
  10. Sep 11, 2010 #9
    an incredibly small fraction, then? hmm so

    well the problem is asking for E based on n = the number of molecules per gram of sample, so since 3.34e22 is the number of molecules per gram of sample, would then E = 1.55e-20?

    or am i somehow supposed to implement the use of a single molecule to find E?
     
  11. Sep 11, 2010 #10

    kuruman

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    That's what it would be equal to.
    I am not sure what you mean by this.
     
  12. Sep 11, 2010 #11
    ohhh ok cool, nevermind i thought there was something to do with the mass of a single molecule...

    anyway and then for part be it says the ratio of such and such:


    would that be E / Kavg of the gas = E / N([3/2]kT) ?

    so E / Kavg = 1.55e-20 / (3.3418e * [3/2] * 1.38e-23 * 302.7) = 7.402e-23?
     
  13. Sep 11, 2010 #12

    kuruman

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    The average energy of a (monatomic) gas molecule is (3/2)kT, so you want E divided by that.
     
  14. Sep 11, 2010 #13
    In the sample it says

    The average kinetic energy of the entire gas is
    K = N(3/2 * kT)

    I tried it both E / that and E over the thing you posted and both answers were wrong? do i have the value of k right = 1.38e-23?
     
  15. Sep 11, 2010 #14

    kuruman

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    Not what you want.

    You did not pay attention to the units. The Boltzmann constant k = 1.38e-23 J/K. You need to convert either kT to calories or E to Joules before you take the ratio.
     
  16. Sep 11, 2010 #15
    ah success! haha thats like the first answer i got completely right XD

    ok one more quick question on a different problem, im pretty sure i know how to do this one only im a tad bit confused, i dont find it difficult enough to post a new thread so ima post it here an see if it'll get answered-

    (a) What is the number of molecules per cubic meter in air at 30C and at a pressure of 1.6 atm (1 atm = 1.01 x 105 Pa)? (b) What is the mass (in grams) of 0.79 m3 of this air? Assume that 75% of the molecules are nitrogen (N2 with a molar mass of 28.00 g/mol) and 25% are oxygen (O2 with a molar mass of 32.00 g/mol).

    ok question is, i know its the ideal gas law i gotta use but its a pV = nRT, im given p & T but not V nor n, am i overlooking something here that's an obvious giveaway for the volume of air or something?
     
  17. Sep 12, 2010 #16

    kuruman

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    Since you are looking for N, use pV = NkT. You know everything else in that expression. You are overlooking that "number of molecules per cubic meter" is another way of saying "number of molecules in one cubic meter."
     
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