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I Heat capacity at constant volume and internal energy

  1. Apr 27, 2016 #1
    I am wondering if equation
    $$C_v=(\frac{∂U}{∂V})_T$$
    applies only to ideal gases or applies generally for any other system?

    The second question I have is can we use the following relation:
    $$dU=nC_vdT$$
    in processes that are non isochoric (that is for processes where volume is not constant)?

    I ask this because in my head this relation should apply only to isochoric processes. But when calculating the work done in adiabatic process for the ideal gas I get the same result if I calculate the integral by using Poisson's relation ##PV^γ=const.## and if I use the above mentioned relation ##dU=nC_vdT##

    $$W=∫pdV=nC_v(T_1-T_2)$$

    So I am confused whether I can use ##dU=nC_vdT## just anytime I need it (and not only for isochoric process) and for any type of system (and not only for ideal gases).
     
    Last edited: Apr 27, 2016
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  3. Apr 27, 2016 #2

    BvU

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    And here's me thinking that ##c_v## is for constant volume processes. Where did you find this equation ?
     
  4. Apr 27, 2016 #3
    Ah sorry, I used the word "isobaric" where I really meant to use "isochoric". Lapsus :(
    I have edited the original post to fix this.
     
  5. Apr 27, 2016 #4

    BvU

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    Generally that's not a good idea: it often makes the thread intractable for others -- and it seduces some my colleagues to fully quote each and every post (because others can't modify the quote any more). That makes threads unbearably repetitious and long.

    So isochoric is with constant volume. Means ##
    C_v=(\frac{∂U}{∂V})_T## can't be right (##\partial V = 0##). Check here for the definition you intended and rephrase in a new post. (or else I still have to quote your original post #1 from a saved copy :smile: ).
     
  6. Apr 27, 2016 #5
    Ahhhh that's another mistake I made, what I really meant was to write:

    $$C_v=(\frac{∂U}{∂T})_V$$
    instead of:
    $$C_v=(\frac{∂U}{∂V})_T$$
    in the original post... I guess I was too tired when typing all this :(

    Should I make a new thread with all fixed typos and delete this one so to stop all this confusion? Please advise :)
     
  7. Apr 27, 2016 #6

    BvU

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    You're fine as it is. So:
    First question: ##
    C_v=(\frac{∂U}{∂T})_V## is general.

    Second question: Therefore: No. Only for isochoric. If the volume changes, the expanding fluid does pV work (or pV work is done on the contracting fluid). That brings in other terms.

    In the adiabatic case you have dQ = 0 and the other terms lead to this http://vortex.nsstc.uah.edu/mips/personnel/kevin/thermo/Chap-3ppt.pdf [Broken] relation

    In summary:
    Isochoric only and for all fluids
     
    Last edited by a moderator: May 7, 2017
  8. Apr 27, 2016 #7
    Can you please elaborate a bit on how is it possible that I get the work done in adiabatic expansion process (δQ=0) for the ideal gas to be the same if I use these two different ways to calculate it:

    1st way by using Poisson's relation:
    $$W=∫pdV$$
    From the Poisson equation we have ##p_iV_i^γ=p_fV_f^γ=pV^γ=const## which gives ##p=\frac{p_iV_i^γ}{V^γ}## which I will put into this integral:
    $$W=∫p_iV_i^γ\frac{dV}{V^γ}=p_iV_i^γ∫\frac{dV}{V^γ}=p_iV_i^γ\frac{V_f^{1-γ}-V_i^{1-γ}}{1-γ}=\frac{p_iV_i^γ}{γ-1}(V_i^{1-γ}-V_f^{1-γ})$$
    Now I will use the fact that ##p_iV_i^γV_i^{1-γ}=p_iV_i=nRT_i## and similarly ##p_iV_i^γV_f^{1-γ}=p_fV_f^γV_f^{1-γ}=p_fV_f=nRT_f##
    then we have
    $$W=\frac{1}{γ-1}(p_iV_i - p_fV_f)=\frac{nR}{γ-1}(T_i - T_f)$$

    and since ##\frac{nR}{γ-1}=\frac{nR}{\frac{C_p-C_v}{C_v}}=\frac{nRC_v}{R}=nC_v## we finally get the end result as:
    $$W=nC_v(T_i - T_f)$$

    2nd way by using ##dU=nC_vdT##:
    $$dU=-δW$$
    $$W=-∫dU=-∫nC_vdT=-nC_v(T_f-T_i)=nC_v(T_i-T_f)$$

    (where ##T_f## and ##T_i## are temperatures of the final and initial state)
    So the work calculated the 2nd way (which is clearly the easier way) is the same as work calculated by the first way when using Poisson's relations. But adiabatic process doesn't need to have constant volume. Is this just a lucky coincidence for the ideal gas or...?
     
  9. Apr 27, 2016 #8

    BvU

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    Over my head somewhat and it's late here. Perhaps @Chestermiller can come to the rescue here ?

    But at first glance you fall into an old trap using cv for non-isochoric (implicit in the word expansion!)
     
  10. Apr 27, 2016 #9
    Yes, I used it on purpose for non-isochoric process here and still got the correct result which lead me to wonder whether ##dU=nC_vdT## can be used universally.
     
  11. Apr 27, 2016 #10
    Not universally but for an ideal gas only.
    The change in internal energy of an ideal gas is given by that formula for any process though. It does not have to be isochoric.

    And is easy to see this. The internal energy is a characteristic of the state of the system. If you have two states 1 and 2, the difference in internal energy is independent of the path you go from 1 to 2. It depends only on the two states. So you know that for an isochoric process from 1 to 2, the work is zero and the change in energy is equal to the isochoric heat transfer so it is given by that formula with Cv. And as it does not depend on the path, will be the same for any other transformation (process) going from 1 to 2.
    The same is not so for heat exchange, which depends on the path. So for heat you will have different formulas, depending on the process.
     
  12. Apr 28, 2016 #11

    BvU

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    nasu: thanks for correcting my oversight. Too much ##c_p## and ##dH##, so I missed out on ##dU## properties.
     
  13. Apr 28, 2016 #12
    For a general material of constant composition and with no phase change, the effect of temperature and volume on internal energy is given by:
    $$dU=C_vdT+\left[T\left(\frac{\partial P}{\partial T}\right)_V-P\right]dV$$
    For an ideal gas, this reduces to ##dU=C_vdT##. So, for an ideal gas over any process path, ##\Delta U=\int C_vdT##, but ##Q=\int C_vdT## only for a constant volume path. Measuring Q over a constant volume path is how we can experimentally determine Cv.
     
    Last edited: Apr 28, 2016
  14. Apr 28, 2016 #13

    vanhees71

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    Well, you have
    $$\mathrm{d} U=T \mathrm{d} S-p \mathrm{d} V,$$
    and thus for ##\mathrm{d} V=0##
    $$C_V=T \left (\frac{\partial S}{\partial T} \right)_V=\left (\frac{\partial U}{\partial T} \right )_V.$$
    For constant pressure you Legendre transform to the enthalpy,
    $$H=U+p V.$$
    Then you have
    $$\mathrm{d} H = T \mathrm{d} S + V \mathrm{d} p,$$
    and thus
    $$C_p=T \left (\frac{\partial S}{\partial T} \right)_{p} = \left (\frac{\partial H}{\partial T} \right )_p.$$
     
  15. Apr 29, 2016 #14
    Thanks for the explanation. However something is still bothering me. Let's say we have ideal gas that goes from state 1 (##p_1, V_1, T_1##) to state 2 (##p_2, V_2, T_2##) in an adiabatic expansion process (eg. like the one in the Carnot cycle). Since this process is expansion we will have ##V_1 ≠ V_2##.

    If I understood you right, you are saying that the reason we can use ##dU=C_vdT## to calculate ##\Delta U=\int C_vdT## for any process of an ideal gas is because: the ##ΔU## is independent of the path between states 1 and 2 and thus we can calculate the ##ΔU## as if we were going between these states along the iso-choric path.
    If this interpretation is correct then my concern is following: in adiabatic expansion that I mentioned above, there will be ##V_1 ≠ V_2## and because of that there exists no iso-choric process that can bring the system from state 1 to 2 because states have different volumes. If no constant volume process can be used between these states then how can we calculate ##\Delta U## that way?
     
  16. Apr 29, 2016 #15
    Another thing is not quite clear to me... I am trying to get to the equation ##dU=C_vdT## in the following way (assuming we are working with ideal gas):

    $$dU = \left(\frac{\partial U}{\partial T}\right)_VdT + \left(\frac{\partial U}{\partial V}\right)_TdV$$
    since for ideal gas ##\left(\frac{\partial U}{\partial V}\right)_T=0## we have will have just ##dU = \left(\frac{\partial U}{\partial T}\right)_VdT##
    and combining with the 1st law ##dU = δQ - pdV = nC_vdT - pdV## we get:
    $$dU = \left(\frac{\partial U}{\partial T}\right)_VdT = nC_vdT - pdV$$
    so if we have an adiabatic process with dV≠0 then how is this equation reduced to just: ##dU=C_vdT##? Where did the ##pdV## term go if ##dV≠0##?
     
  17. Apr 29, 2016 #16
    Yes, this is a good point and I should have been more specific.
    You cannot go directly on an isochoric process but you can go in two steps. One isochoric and one isothermal.
    (P1,V1,T1)->(P',V1,T2)->(P2,V2,T2)
    In the second process the change in energy is zero. Like in the figure.
    Drawing the diagram helps. Try to apply the same idea for an isobaric transformation.
    125q7mr.jpg
     
  18. Apr 29, 2016 #17
    For a non-constant volume process, ##Q\neq \int{nC_vdT}##. For an adiabatic process of an ideal gas with ##\Delta V \neq 0##, $$\Delta U=\int{nc_VdT}=-\int{p_I}dV$$where ##p_{I}## represents the externally applied pressure at the interface with the surroundings. If the path is reversible, then, along the process path, ##dU=nc_VdT=-p_IdV=-pdV## and ##\Delta U=\int{nc_VdT}=-\int{pdV}##.

    Since U is a thermodynamic function of state, it depends only on the initial and final states, and not the process path between these states. So stop focusing on process paths. They are irrelevant. The only kind of process path for which you should be using a differential dU is along a reversible path, which is comprised of a continuous sequence of equilibrium states. For an irreversible path, you're stuck with using ##\Delta U=\int_{T_1}^{T_2}nC_v(T')dT'##, where T1 and T2 are the temperatures at the two end states, but where T' is a dummy variable of integration that does not relate in any way to the (usually non-uniform) temperatures of the system along the irreversible process path.

    The equation I presented in post #12 was derived on the basis of differential changes between closely neighboring equilibrium states of a material. This equation can be integrated between non-neighboring pairs of thermodynamic equilibrium states.

    For more details on all this, see my Physics Forums Insights article, https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/
     
    Last edited: Apr 29, 2016
  19. Apr 29, 2016 #18
    @misko What you need to realize is that the equilibrium internal energy U(T,V) (or equivalently, U(T,P)) is a physical property of the material, independent of any process. How this works is: You tell me the values of T and V, and I will tell you the value of U, irrespective of how the material got to the state T,V. So, if we want the change in internal energy between the coordinates T1, V1 and T2, V2, we can write $$\Delta U=U(T_2, V_2)-U(T_1,V_1)$$. We can write this as:$$\Delta U=(U(T_2, V_1)-U(T_1,V_1))+(U(T_2,V_2)-U(T_2,V_1))$$ But, for an ideal gas, ##(U(T_2,V_2)-U(T_2,V_1))=0##. So, for an ideal gas, $$\Delta U=(U(T_2, V_1)-U(T_1,V_1))$$. In the same way, we can also show that $$\Delta U=(U(T_2, V_2)-U(T_1,V_2))$$

    Chet
     
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