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Calculate ΔU. n is constant. V, T and p change

  1. Sep 15, 2014 #1
    1. The problem statement, all variables and given/known data
    Calculate ΔU when 1.00 mol of H2 goes from 1.00 atm, 10.0 L, and 295 K to 0.793 atm, 15.0 L, and 350 K.


    2. Relevant equations
    ΔU = q + w
    q=mcΔT


    3. The attempt at a solution
    From the moles of H2 we can get the mass of H2 and use q=mcΔT.
    1.00 mol H2=(2.0158 g H2/1 mol H2)=2.0158 g H2
    Thus:
    q=mcΔT
    q=(2.0158 g)(14.314J/mol*K)(55K)
    q=1586 J

    The answer section in my book says ΔU = 1590 J. My answer is close enough but what about w? The problem doesn't explicitly say the gas is subject to a constant external pressure so I can't use w=pextΔV.
     
    Last edited by a moderator: Sep 15, 2014
  2. jcsd
  3. Sep 16, 2014 #2
    Does your book say that ΔU = 1590 J, or q = 1590 J?

    ΔU ≠q

    =================================

    Work is calculated with this integral:

    [itex]W = \int_{V_1}^{V_2} p\,dv[/itex]

    Which means you need to find a formula for pressure, because it is not constant.
     
    Last edited: Sep 16, 2014
  4. Sep 16, 2014 #3

    Borek

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    Staff: Mentor

    I would assume answer in the book is identical to your answer - it is just rounded to follow the number of significant digits given in the problem. All numbers are given to 3 sf, your answer gives 4.
     
  5. Sep 16, 2014 #4
    This is follow-up to what TheOrange said. In this problem, q is not equal to ΔU. The equation you wrote for q should really be the equation for ΔU. In the ideal gas region, ΔU=mCvΔT always, irrespective of whether the volume is constant, since Cv is defined by the equation:

    [tex]mC_v=\frac{∂U}{∂T}[/tex]

    Chet
     
  6. Sep 17, 2014 #5
    Look at an ideal gas, the formula below may be helpful as well:

    [itex]P*V^n = constant[/itex]
     
  7. Sep 17, 2014 #6
    He was only asked to determine ΔU, not the work or the heat.

    Chet
     
  8. Sep 17, 2014 #7
    [itex]\Delta U = Q - W[/itex]

    Which btw he already posted as a relevant equation...
     
  9. Sep 17, 2014 #8
    He may have posted this as a relevant equation, but it is obviously not needed for this problem. For any ideal gas,
    [tex]ΔU=mC_vΔT[/tex]
    and, in this problem,
    m=1 mole
    Cv=28.836 J/(mole K)
    ΔT=55 K

    The clear purpose of this exercise was to test the understanding of the student to determine whether he was aware that, for an ideal gas, the change in internal energy could be calculated from ##ΔU=mC_vΔT##, irrespective of the changes in pressure and volume.

    In any case, there certainly isn't enough information provided to determine either Q or W, particularly since no indication is given of whether the process is reversible or irreversible.

    In thermo, it is very important to recognize that the internal energy is a physical property of the material.

    Chet
     
  10. Sep 18, 2014 #9
    He asked himself, "My answer is close enough but what about w? The problem doesn't explicitly say the gas is subject to a constant external pressure so I can't use w=pextΔV."

    So I was just trying to help by saying that he needs to find a formula for P (since it isn't constant), and that W is an integral.
     
  11. Sep 18, 2014 #10
    It was already pointed out by Borek in post #3 with regard to the comparison between the book and the OP's calculation, that the OP was dealing with a significant figures issue.

    The work W is always equal to ##\int{P_{ext}dV}##, irrespective of whether the process is reversible or irreversible. However, we can be certain that Pext=P (with P determined from the equation of state for the material) only if the process is reversible.

    Chet
     
  12. Sep 18, 2014 #11
    Chet, seriously what are you talking about. What Borek said had nothing to do with what I said. The formula for work using [itex]\Delta V[/itex] does not work in this case because P is not constant. Therefore I was letting him know he needs to use the integral form and find an equation for P. Which can be found using the ideal gas formula. I also said that it MIGHT be helpful.

    I really don't see what the problem is. To avoid spamming this topic even more maybe you should just message me, if this is SO important to you.
     
  13. Sep 18, 2014 #12
    It is important to me, because I didn't want the OP to get confused. But I like your idea about discussing this in private messages so that we can reach a consensus. Then we can report back jointly to the thread. Is this agreeable to you?

    Chet
     
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