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Question regarding thermodynamics

  1. Nov 26, 2013 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    Why dU=nCdT true for all processes?
     
  2. jcsd
  3. Nov 26, 2013 #2
  4. Nov 26, 2013 #3

    rude man

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    It is only true for an ideal gas, and only for a quasi-static process.

    And C should read cv, the specific heat capacity at constant volume.
    So it's dU = ncvdT.
     
  5. Nov 27, 2013 #4

    utkarshakash

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    Is ΔQ=nCpΔT true for all processes?
     
  6. Nov 27, 2013 #5
    No...only for isobaric process.
     
  7. Nov 27, 2013 #6

    utkarshakash

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    Is it because the equation involves Cp ? If it is, then returning to my original question, change in internal energy also involves Cv but nevertheless, it remains true for all processes whether or not the volume is constant (assuming ideal gas). How can this be justified?
     
  8. Nov 27, 2013 #7
    Yes...There are infinite number of molar specific heat capacities.The two important and also easy to calculate are CP and CV

    You can understand this in three steps.

    1. ΔU = (f/2)nRΔT .This applies to all kinds of processes involving ideal gases.

    2. Next consider an isochoric process(constant volume process) .

    Using First Law of Thermodynamics, ΔQ = ΔU + ΔW

    Now, ΔQ = nCvΔT ,ΔU = (f/2)nRΔT and ΔW = 0

    So,we have nCvΔT = (f/2)nRΔT

    or, Cv = (f/2)R .i.e molar heat capacity at constant volume for an idela gas is a constant .

    3. Now come back to isobaric process .

    As we have noted in point 1 , ΔU = (f/2)nRΔT , in an isobaric process for an ideal gas .But Cv for the gas (even though the gas is undergoing a constant pressure process) is equal to (f/2)R .

    So we have ΔU = nCvΔT.

    Thus ,we see that this relation ΔU = nCvΔT applies to all kinds of processes involving an ideal gas ,just like ΔU = (f/2)nRΔT holds for any process involving an ideal gas.
     
  9. Nov 27, 2013 #8

    rude man

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    Definition of an ideal gas:
    1. pV = nRT
    2. ∂U/∂p at constant T = 0.

    From the above, you can show that ∂U/∂V at constant T also = 0.

    So U is a function of T only.

    First law: dU = δQ - pdV

    and CV = δQ/dT at constant volume by definition;
    but at constant volume, dV = 0, so

    Cv = ∂U/∂T at constant V.

    But since U = U(T) only, and not a function of V, then the partial derivative can be replaced by the total derivative and we wind up with

    dU = CVdT for all quasi-static processes of an ideal gas.
     
  10. Nov 30, 2013 #9

    CAF123

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    I have a more conceptual question related to this; Since dU is a state variable, a change in dU of a material between two points on a PV diagram is independent of the path taken. So, given this, why can we not define, for example, a constant pressure process between these two end points and define dU = CPdT?
     
  11. Nov 30, 2013 #10
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    Last edited: Nov 30, 2013
  12. Nov 30, 2013 #11

    rude man

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    Cp = δQ/dT at constant p, not dU/dT at constant p.

    Cp = δQ/dT|p by definition; but
    δQ = dU + p dV
    but if p is constant, V is not, otherwise there is no change in state;

    so Cp ≠ dU/dT|p.
     
  13. Nov 30, 2013 #12

    CAF123

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    Hi rude man. I understand that, but I am trying to understand how it agrees with the fact that internal energy is a state variable of the substance. As far as I understand, the change in internal energy of a substance between two points on a PV diagram does not depend on the path taken. So why not consider a constant pressure process and write dU = cpdT?

    Alternatively, even for a non constant volume process dU = cvdT still holds. How does your argument take this into consideration?
    Thanks,
     
  14. Nov 30, 2013 #13

    rude man

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    I think I did my best to explain. Maybe someone else can do a better job.
     
  15. Nov 30, 2013 #14
    dU and dH are given as follows:
    [tex]dU=TdS-PdV[/tex]
    [tex]dH=TdS+VdP[/tex]
    For dH, we can also write:
    [tex]dH=T(\frac{\partial S}{\partial T}dT+\frac{\partial S}{\partial P}dP)+VdP=T\frac{\partial S}{\partial T}dT+(T\frac{\partial S}{\partial P}+V)dP[/tex]
    The heat capacity at constant pressure is defined as:
    [tex]C_p\equiv\frac{\partial H}{\partial T}[/tex]
    Therefore, from the previous equation,
    [tex]dH=C_pdT+(T\frac{\partial S}{\partial P}+V)dP[/tex]
    together with,
    [tex]C_p=T(\frac{\partial S}{\partial T})_P[/tex]
    For an ideal gas, it can be shown that:
    [tex](T\frac{\partial S}{\partial P}+V)=0[/tex]
    Therefore, for an ideal gas,
    [tex]dH=C_pdT[/tex]
    This is independent of pressure.

    Now, for dU we can also write:
    [tex]dU=T(\frac{\partial S}{\partial T}dT+\frac{\partial S}{\partial V}dV)-PdV=T\frac{\partial S}{\partial T}dT+(T\frac{\partial S}{\partial V}-P)dV[/tex]
    The heat capacity at constant volume is defined as:
    [tex]C_v\equiv(\frac{\partial U}{\partial T})_V[/tex]
    Therefore, from the previous equation,
    [tex]dU=C_vdT+(T\frac{\partial S}{\partial V}-P)dV[/tex]
    together with,
    [tex]C_v=T(\frac{\partial S}{\partial T})_V[/tex]
    For an ideal gas, it can be shown that:
    [tex](T\frac{\partial S}{\partial V}-P)=0[/tex]
    Therefore, for an ideal gas,
    [tex]dU=C_vdT[/tex]
     
    Last edited: Nov 30, 2013
  16. Nov 30, 2013 #15

    utkarshakash

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    Nicely explained. :cool:
     
  17. Nov 30, 2013 #16
    I made several corrections to my original posting #14. Please disregard what I wrote previously. The posting is now correct. Sorry for any confusion I may have caused.

    Chet
     
  18. Dec 5, 2013 #17
    The first Law, which is applicable to all processes says: Q = ΔU + W. When only mechanical work (PdV work) is involved we write the first law as:Q = ΔU + PdV. For a constant volume process dV = 0. Therefore, Q = ΔU = nCVdT.

    It is true that U is a state function and has a unique value for ΔU connecting two equilibrium states A, B of a system. If A and B are conneted by a const volume process, then A and B cannot be connected by a const pressure process. A constant pressure process takes the system from A to state C (≠B). ΔUAB≠ ΔUAC.

    Therefore, ΔU for a constant volume process is not equal to ΔU for a constant pressure process.

    However, we can take the system from A to C by a constant pressure process and from C we can take it to state B by one more process (or step). If the system is an ideal gas, then if B and C have the same temperature, that is if B and C lie on a reversible isotherm then ΔUAB= ΔUAC, since U of an ideal gas is a function only of temperature and is not a function of volume or pressure.

    Thus for an ideal gas: QV = nCVdT = ΔU
    QP = nCVdT + PdV = ΔU + PdV.
     
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