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Thermodynamics, ideal gas relation

  1. Mar 17, 2012 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Find the relation [itex]P=P(V)[/itex] for a transformation [itex]dQ=0[/itex] in an ideal gas ([itex]PV=nRT[/itex] and [itex]U=CnRT[/itex]).


    2. Relevant equations
    [itex]dU=dQ-PdV[/itex].


    3. The attempt at a solution
    If I assume that C and R are constant I get [itex]dU=CR \left [ \frac{\partial (nT)}{\partial n} dn + \frac{\partial (nT)}{\partial T } dT \right ] =-PdV[/itex].
    If I assume that n does not depend on T, this simplifies to [itex]dU=CR(Tdn+ndT )=-PdV[/itex]. If I assume that n is constant, dn=0 and so [itex]-PdV=CRndT[/itex]. But since the pressure can depend on the volume, I cannot just integrate this equation. I don't know how to find P(V). Also I think I assumed too many things that weren't stated in the problem statement. Any idea on how to proceed?
     
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  3. Mar 18, 2012 #2

    rude man

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    EDITed out. This is deriving the fundamental p-V relationship of an adiabatic process in an ideal gas ...
     
    Last edited: Mar 18, 2012
  4. Mar 18, 2012 #3
    Whenever, you come across such questions its best to assume that PV^x or PT^x is constant.(valid only when amount of gas i.e effective number of moles are not changing)

    Now you have to find what the value of x and constant are.

    Apply two laws.
    Ideal gas law and 1st law of thermodynamics

    However, since you have made a different approach(and you have dine quiet well) I would like you to continue with that.

    You have used the 1st law of Thermodynamic.Now use the ideal gas law to substitute dT.
     
  5. Mar 18, 2012 #4

    fluidistic

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    Thanks guys. Yeah I'm going to stick with my way of doing it. So you suggest me to use the relation [itex]PV=nRT[/itex] to get [itex]dT[/itex] and then replace in [itex]-PdV=CRndT[/itex].
    So I get [itex]dT=\left ( \frac{\partial T}{\partial P} \right ) dP+\left ( \frac{\partial T}{\partial V} \right ) dV[/itex]. Replacing into [itex]-PdV=CRndT[/itex] does not help much I think. I'm wondering if this is the way to go.
     
  6. Mar 18, 2012 #5

    rude man

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    This is derived in every calculus-level physics and thermodynamics textbook you'll ever run into, guaranteed.
     
  7. Mar 19, 2012 #6
    The ideal gas equation is PV=nRT

    Differentiating we get PdV+VdP =nRdT

    Sunstitute dT now.

    Don't use partial derivatives
     
  8. Mar 19, 2012 #7

    fluidistic

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    Thanks guys, I'd rather do it myself with your help :)
    So [itex]dT=\frac{PdV+VdP}{R}[/itex].
    [itex]\Rightarrow c(PdV+VdP)=-PdV \Rightarrow -P(1+c)dV=cVdP \Rightarrow -(1+c) \ln V =c \ln P + K \Rightarrow P= \frac{(V^{-1-c})^{1/c}}{B}[/itex] where B is a constant.
    I have never seen that before, so I'm guessing I made error(s)...
     
  9. Mar 19, 2012 #8
    You have got it correctly.:-)


    Just get V on the other side and substitute c.
    You will get PV^x=constant where x is gamma i.e C(P)/C(V).
    The constant(1/B) will depend on initial conditions
    :-)

    p.s. (Avoiding using symbol B as constant.People may confuse it with Bulk Modulus in Thermodynamics.Strangely its better to rather use x,y,z :P
     
  10. Mar 19, 2012 #9

    rude man

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    Hate to crash the party, but that is incorrect. C is a heat capacity but as emailanmol alludes to in his last post we need a ratio of heat capacities, not just one heat capacity. You have been dealing with just one heat capacity so far if I'm not mistaken (which I often am ... ) :rolleyes:
     
  11. Mar 19, 2012 #10

    Lol Rudeman :-)


    Actually C is not heat capacity here.

    From what i can see, C is degrees of freedom/2.
     
    Last edited: Mar 19, 2012
  12. Mar 19, 2012 #11

    rude man

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    OK, [itex]U=CnRT[/itex]). That's from the original problem statement. So C here is actually CV/nR = cV/R, CV is total heat capacity at constant volume, cV is molar specific heat at constant volume, and n= no. of moles of the gas.

    The relation dU = CVdT is a direct consequence of the definition of an ideal gas.

    My point is he's missing another heat capacity, that at constant pressure. You obviously know the correct formula but he's not there himself. He has to derive it whereas you & I know it from our textbooks and/or instructors.


    BTW I have problems with that statement to begin with. U = CnRT assumes validity down to absolute zero which is by no means a fait accompli at the OP's stage of learning (or mine, for that matter). It's a lot safer to just say dU = CnRdT or as I prefer it, dU = CVdT.
     
  13. Mar 19, 2012 #12
    As you said C= C(V)/R and C(V)/R is degrees of freedom/2.

    Your other missing heat capacity is the term c+1 in power of V. ( what is (c+1)R ?)


    As for the last point, i think the OP made a typo by writing U=CnRT .
    It should be dU=CnR(dT).

    He has used the correct formula in this post

    And in his final answer.

    His answer is perfect . :-)

    As i mentioned the only thing he needs to work on is to avoid using standard variable as different ones .(which is causing this whole confusion over c)
    But if he does he should declare them in advance as he did for B. :-)
     
    Last edited: Mar 19, 2012
  14. Mar 19, 2012 #13

    fluidistic

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    I didn't make a typo when writing [itex]U=CnRT[/itex].
    I reached the formula [itex]-PdV=CRndT[/itex] because [itex]dU=dQ-PdV[/itex] and [itex]U=CnRT[/itex]. I calculated [itex]dU[/itex] assuming n, R and C as constants, yielding [itex]dU=CnRdT[/itex].

    Anyway I did what you said (try to get PV in one side), reaching [itex]PV^{1+1/c}=K_2[/itex] where [itex]K_2[/itex] is a constant.
    I hope this is right. Actually I prefer to derive [itex]PV^\gamma = \text{constant}[/itex] rather than getting a "useless" P(V) as they asked me. Well I should get both. :biggrin:
     
  15. Mar 19, 2012 #14
    Hey. U =CnRT is not right.

    dU =CnRdT is right.

    U=CnRT+ U(0)

    Which goes away on differentiating :-)

    Also you do realise that 1+c/c is gamma.Right ?

    ( there is a typo in your previous post in power of V where you have written 1+1/c)
     
  16. Mar 24, 2012 #15

    fluidistic

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    Sorry for being so late (too busy with other courses, it's hard not to fall behind in all courses!).
    I can see U=CnRT in wikipedia: http://en.wikipedia.org/wiki/Ideal_gas, are you sure it's wrong? Basically the problem consider an ideal gas whose internal energy tends to 0 when the temperature tends to 0. I don't know if it's relevant for this problem though, but I do think the equation is right.
    Do you mean [itex]\gamma=\frac {1+c}{c}[/itex]? In which case it's worth 1/c+1 which is what I found. I don't see my typo in post #13.
     
  17. Mar 24, 2012 #16
    Taking U as CnRT + U(0) is a good practice, as defining energy requires us to define a reference point.No one knows what will happen at 0K :-)

    Practically and Logically it makes no difference to skip writing U(0) but it is a good practice to write it. :-)

    Yes
    y(gamma)=(1/c) +1


    The typo (which isn't there.Sorry !) in your post is that
    you wrote PV^1+1/c=K2

    I thought you were writing (1+1)/c (which is really stupid of me :P).
    It is actually fine :-)


     
  18. Mar 24, 2012 #17

    fluidistic

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    Okay, thank you very much. So basically the problem is solved, right? :smile:
     
  19. Mar 24, 2012 #18
    Yes, it is :-)
     
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