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Thermal physics

  1. Apr 25, 2008 #1
    [SOLVED] thermal physics

    1. The problem statement, all variables and given/known data
    the differential of enthalpy is given by dH=TdS-VdP


    2. Relevant equations
    Beginning with the above relationship, find the equation for the differential of the entropy and then show that (d(1/T)/dp)|H=(d(V/T)/dH)|p


    3. The attempt at a solution

    this is what i got so far...
    i rearranged dH=TdS-VdP to make S the subject which gives dS=dH/T+Vdp/T
    then since S=S(H,p) and using the definition of the differential of S;
    dS=(dS/dH)|p*dH+(dS/dp)|H*dp
    i got then 1/T=(dS/dH)|p & V/T=(dS/dp)|H
    i dont know what to do next and to how to exactly reach the final term of (d(1/T)/dp)|H=(d(V/T)/dH)|p

    please help , kind regards
     
  2. jcsd
  3. Apr 25, 2008 #2

    Hootenanny

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    Welcome to PF billybob,

    You're on the right lines and almost there. What do you know about the mixed second partial derivatives of an exact differential?
     
  4. Apr 25, 2008 #3
    exactness condition
    i know that they are equal, i thought i would end up somewhere along those lines so..

    dS=(dS/dH)|p*dH+(dS/dp)|H*dp

    =M(H,p)dH+N(H,p)dp

    dM/dp=dN/dH

    i see where its getting next bit algebra is getting tricky..

    do i cheat and just sub (1/T) for M and and V/T for N?

    i dont understand how to show this mathematically nicely
     
  5. Apr 25, 2008 #4

    Hootenanny

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    Correct.
    I'm not quite sure what you doing here, but you know that,

    [tex]\left(\frac{\partial S}{\partial H}\right)_p =\frac{1}{T}[/tex]


    [tex]\left(\frac{\partial S}{\partial P}\right)_H =\frac{V}{T}[/tex]

    And you know that,

    [tex]\left[\frac{\partial}{\partial P}\left(\frac{\partial S}{\partial H}\right)_p\right]_H = \left[\frac{\partial}{\partial H}\left(\frac{\partial S}{\partial P}\right)_H\right]_P[/tex]

    Can you see what you need to do next (it's a really simple step)?
     
    Last edited: Apr 25, 2008
  6. Apr 25, 2008 #5
    subbing in becames..
    (d/dp)(1/T)=(d/dH)(V/T)
    i cant see how the |H and |p get switched around
     
  7. Apr 25, 2008 #6

    Hootenanny

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    Correct. :approve:
    Sorry my bad, I left out the subscripts of what I'm holding constant in my previous post. Does it make more sense now?
     
  8. Apr 25, 2008 #7
    Yes! thanks you so much appreciate it heaps

    going to attempt part (b) of the question ill post here if i having difficulties

    thanks again!
     
  9. Apr 25, 2008 #8

    Hootenanny

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    Your welcome :smile:
     
  10. Apr 27, 2008 #9
    part (b)

    now for part (b) of this question

    Using the equation of state for an ideal gas and the definition of enthalpy (H=U+pV), express 1/T and V/T as function of p and H and show that the relationship in part (a) is valid for an ideal gas.

    trying to understand what this question means..

    i know pV=nRT, do i sub that in enthalpy equation which becomes H=U+nRT?

    want 1/T(p,H) and V/T(p,H)

    i really don't know where to start for this question and how to approach it.

    some hints will be much appreciated;

    thanks in advance.
     
    Last edited: Apr 27, 2008
  11. Apr 27, 2008 #10

    Hootenanny

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    So what you want is

    [tex]\frac{1}{T}=f(p,H)\hspace{2cm}\frac{V}{T}=F(p,h)[/tex]

    With respect to the definition of the enthalpy, bearing in mind that we are dealing with an ideal gas (no potential energy), can you write U as a function of T?
     
  12. Apr 27, 2008 #11
    i thought U=0 like you said above, since its an ideal gas which makes H=pV;
    i am sorry i don't understand writing U as a function of T since U is 0;

    so V=H/p

    pV=nRT => p(H/p)=nRT => T=H/nR

    that right?

    sorry i am completely puzzled by this
     
  13. Apr 27, 2008 #12

    Hootenanny

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    Not quite. U is the internal energy, which roughly speaking is the sum of the potential energies and kinetic energies of the gas. Now as I said previously, since the gas is ideal, there is no potential energy; but there is still kinetic energy. Now, how is the kinetic energy of an ideal gas related to it's temperature? (HINT: Equipartition theorem).
     
  14. Apr 27, 2008 #13
    the relationship is;
    change is kinetic energy means a change in temperature (proportional to each other).

    equipartition theorem = 3/2RT

    write U as a function of T is is U=3/2RT or U=C*T (where C is a constant)

    so H=U+pV

    H=CT+pV
    that correct? and then make V/T and 1/T the subject?
     
  15. Apr 27, 2008 #14

    Hootenanny

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    Are you sure it's R? Apart from that it looks good.
     
  16. Apr 27, 2008 #15
    its 3/2kT sorry;

    ok so rearranging to make 1/T and V/T the subject i get
    1/T=1/(H-pV)
    V/T=H/TP - 1/P => for this one im not sure cause V/T is not the clear subject there is a T term in the equation

    and now i should plug it into equation in part (a) then solve?

    i thank you for your continuing help
     
    Last edited: Apr 27, 2008
  17. Apr 27, 2008 #16

    Hootenanny

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    Better :approve:
    I think your missing a factor of C here.
    For this one, I think it would be must easier to use the ideal gas eqaution and simply find V/T as a function of P.
     
  18. Apr 27, 2008 #17
    sorry my algebra is a bit weak

    H=kT+pV where pV=nRT
    H=kT+nRT
    i dont see how this can help
    and here probably alot more worse i know this is wrong but im trying everything.
    H=kT+n^2*R^2*T^2/Vp (i got this from p=nRT/V and V=nRT/p and taking the product of this yeilds n^2*R^2*T^2/Vp)

    this is embarassing

    maths...
     
  19. Apr 27, 2008 #18

    Hootenanny

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    Your second expression need not be a function of H, what I meant was,

    [tex]PV = nRT \Rightarrow \frac{V}{T} = \frac{nR}{P}[/tex]

    Do you follow?
     
  20. Apr 27, 2008 #19
    yes i understand where u got that from but my partial derivative on RHS is [d(V/T)/dH]|p so i need it in terms of H so i can proceed with the partial and proove they are equal
     
  21. Apr 27, 2008 #20

    Hootenanny

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    You're quite right, I forgot that the partial derivative was with respect to H :redface:. So, using the definition of enthalpy and the ideal gas law,

    [tex]H = \left(C+nR\right)T \Rightarrow T = \frac{H}{C+nR}[/tex]

    And,

    [tex]PV = nRT \Rightarrow \frac{V}{T} = \frac{nR}{P}[/tex]

    Now, you should be able to combine the two and hence form an expression for V/T in terms of P and H.
     
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