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Thermodynamics Problem

  1. Mar 1, 2012 #1
    1. The problem statement, all variables and given/known data

    http://img51.imageshack.us/img51/9484/therm.jpg [Broken]

    3. The attempt at a solution

    My main problem is with the second part of the question. Here's my attempt at the first part:

    [itex]p = \rho RT \implies \rho = \frac{p}{RT} = \frac{1010 \times 10^2 \ Pa}{287 \times 300.15 K}[/itex]

    [itex]\therefore p = 1.1724 \ Pa[/itex]

    (R was the specific gas constant for dry air = 287)

    And for the second part:

    If air is modeled as an ideal gas the only internal energy is kinetic and we can use:

    [itex]K = \frac{3}{2} n R T[/itex]

    (where R is the universal gas constant = 8.314)

    But how do we find the number of moles for dry air? We must use the equation n=m/M. Here M is the molar mass (atomic number). But air is not a single gas, it is composed of several different gases. So how do we know its molar mass?? Do we have to use the atomic number of oxygen or nitrogen or carbon? :confused:

    I appreciate any advice. If I've gone wrong anywhere else I would appreciate if somebody lets me know.
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Mar 1, 2012 #2


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    You wanted to determine the density, not the pressure. Do not mix them.

    The specific gas constant = R(universal)/(molar mass).

    Last edited: Mar 1, 2012
  4. Mar 1, 2012 #3

    Andrew Mason

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    As you note, air consists of mainly diatomic molecules. For temperature below 1000K, Cv = 5R/2 where R is the UNIVERSAL gas constant.

    Use [itex]U = nC_vT[/itex] for the internal energy.

  5. Mar 1, 2012 #4
    Thanks it was a typo, I meant to say [itex]\rho = 1.1725 \ Pa[/itex]

    Since we are given that the specific gas constant for dry air, R*=287 J K-1 kg-1, then

    [itex]M= \frac{R}{R^*} = \frac{8.314}{287} = 0.02896 \ g/mol[/itex]

    [itex]n = \frac{m}{M} = \frac{1000}{0.02896}= 34520.1[/itex]

    So, the number of moles is 34520.1?

    What does the symbol Cv represent?

    [itex]C_v = \frac{5 \times 8.314}{2} = 20.78[/itex]

    And why can't we just use Kint=3/2nR*T? (in an ideal gas we only consider the kinetic since molecules interact only by collisions)
  6. Mar 1, 2012 #5

    Andrew Mason

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    Cv is heat capacity at constant volume. You cannot use Cv = 3R/2 because the diatomic molecules have rotational kinetic energy as well as translational kinetic energy resulting in 5 degrees of freedom. (It does not have vibrational kinetic energy at this temperature for reasons having to do with quantum mechanics). So U = n(5/2)RT

  7. Mar 1, 2012 #6

    Andrew Mason

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    ?? Check this.

    ?? Better check this too - units: grams or kg?.

  8. Mar 1, 2012 #7
    Oops. Thank you. I fixed the unit. :redface:

    I just solved for the density ρ (rho).

    ρ = p/RT => (1010 x 102)/ (287 x 300.15) = 1.1725 Kg/m3

    Was that the only problem?


    [itex]M = \frac{R^*}{R}= \frac{8.314 \ J/mol.K}{287 \ J K^{-1} Kg^{-1}} = 0.02896 \ Kg/mol[/itex]

    [itex]n = \frac{m}{M} = \frac{1}{0.02896} = 34.52 \ moles[/itex]

    [itex]K_{int} = \frac{3}{2} \times 34.52 \times 8.314 \times 300.15 = 129214.575 \ J[/itex]

    Is this good?

    I'm sure you are right but we only have the formula with 3/2 in our notes, we don't have the one with the 5/2. I think maybe for simplicity we only consider 3 degrees of freedom (x, y, z).
  9. Mar 1, 2012 #8


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    It would be wrong, as a diatomic molecule has got 5 degrees of freedom.

    You can get the internal energy from Cv, the specific heat capacity at constant volume. The internal energy of an ideal gas is Cv m T.

    Have you got a table for specific heat capacity of air? Like that:

  10. Mar 2, 2012 #9
    So the calculation I did in my last post would be correct for a monatomic molecule? Also what exactly are the other 2 degrees of freedom that a diatomic molecule has?

    So is this calculation correct now for air:

    [itex]U_{int} = \frac{5}{2} \times 34.52 \times 8.314 \times 300.15 = 215357.1 \ J[/itex]

    P.S. Your link says that Cv around 300 K must be 0.718. But 8.314x5/2=20.7! :confused:
  11. Mar 2, 2012 #10


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    Both main components of air, nitrogen and oxygen, form diatomic molecules. They are like dumbbells. Rotation about two axis, normal to the axis of symmetry of the molecule, contributes to the internal energy.

    It is correct with the approximation that the dry air is ideal gas.

    The link gives measured specific heat capacity which is the heat needed to warm up 1 kg gas by 1K, not 1 mol gas, and it is given in kJ/(kgK). F/2 R is the theoretical molar heat capacity.

    Multiply 718 by 1 kg and 300.15 K: It is 215507 J. Very near to your result.

  12. Mar 2, 2012 #11
    Thank you very much for the explanation, that makes perfect sense now.
  13. Mar 3, 2012 #12


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    You are welcome.

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