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Homework Help: Thermodynamics - Ideal Diatomic Gas

  1. Nov 6, 2005 #1
    The question is as follows:
    An ideal diatomic gas for which cv = (3/5)*R, where R is the ideal gas constant, and occupies a volume V= 2 meters^3, at a Pressure Pi= 5 atm and temperature Ti = 20 degrees Celcius (= 293 K). The gas is compressed to a final pressure of Pf = 12 atm. Calculate the final volume Vf, the final temperature Tf, the work done dW, the heat transferred dQ, and the change in Internal energy dU for the following:
    a). a reversible isothermal compression
    b). a reversible adiabatic compression
    i know in the adiabatic process, dQ = 0, and in the isothermal dT = 0. Where do i take it from here? :confused:
  2. jcsd
  3. Nov 6, 2005 #2
    anyone please?
  4. Nov 6, 2005 #3

    Andrew Mason

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    The first one is fairly straightforward. Think PV=nRT. If T is constant, what does that tell youi about PV at all times? That should give you Vf. What is U at intial and final volumes? What is W in terms of P and [itex]\Delta V[/itex]? That should give you W. What is the relationship between U, Q and W? That will give you Q.

    For the second part, it is more complicated. You have to use the adiabatic relationship:

    [tex]PV^\gamma = constant[/tex]
    where [itex]\gamma = C_p/C_v[/itex]

    What is [itex]\gamma[/itex] here?
    What is the expression for W?
    How is W related to Q and U?

  5. Nov 8, 2005 #4
    Please HELP!!!

    :surprised i am stuck with the internal energy for part a) isothermal process.

    here is what i've done so far. since its isothermal:

    P1*V1 = P2*V2​
    and thus,
    V2= (P1/P2)*V1 = 0.8333333 m^3

    For isothermal, Ti = Tf = T = constant = 293.15 K

    W = -nRT*(Integral: Pi to Pf) (dP/P) = -nRT*ln(P2/P1) = nRT*ln(P1/P2)
    thus W = -2.134 * 10^6 J, work done on the system

    Now, for U i am stuck. i keep reading in the book, that for an ideal gas U only depends on T, and they give me this formula.

    U - U_0 = dU = Cv*dT, where Cv is the specific heat at constant volume. but volume is not constant in this problem, and also, this leads to an integral with dT and for isothermal dT = 0
    If this is true, then the integral yields a constant U

    and that means U_final = U_initial, but that still doesnt tell me what U is.

    Anything further? where am i going wrong?
  6. Nov 8, 2005 #5
    please help, need to turn this in tomorrow and im stuck
  7. Nov 8, 2005 #6

    Andrew Mason

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    U is the internal energy due to kinetic energy of the molecules. If you keep the volume constant and increase the temperature, the gas does no work. So the increase in internal energy is equal to the heat input. Since the heat flow is just the heat capacity of the gas at constant volume multiplied by the change in temperature, [itex]dU = nC_vdT => U = \int dU = \int nC_vdt = nC_vT[/itex].

    So [itex]U = nC_vT[/itex] (T in K).

    Last edited: Nov 8, 2005
  8. Nov 9, 2005 #7
    understood now, thanks very much AM
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