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Internal Energy of virial expansion

  1. May 25, 2017 #1
    Hello,
    I have some trouble understanding the virial expansion of the ideal gas.

    1. The problem statement, all variables and given/known data

    I have given the state equation:
    $$ pV = N k_b T \left(1+\frac{A\left(T\right)}{V}\right) $$


    2. Relevant equations

    and a hint how to calculate the caloric equation of state $$ \left(\frac{\partial U}{\partial V}\right)_T = T^2 \left(\frac{\partial}{\partial T}\left(\frac{p}{T}\right)\right)_V $$


    3. The attempt at a solution

    I calculated ## \frac{\partial}{\partial T} \left(\frac{p}{T}\right)_V = \frac{N k_b}{V^2} \left(\frac{\partial A\left(T\right)}{\partial T}\right)## first. So using the hint $$ U = \int_0^V \frac{N k_b}{V^2} T^2 \frac{\partial A\left(T\right)}{\partial T} dV = -N k_b T^2 \frac{\partial A\left(T\right)}{\partial T} \frac{1}{V}$$
    If my calculations are correct, why is my internal energy negative? Unfortenately, I don't know which sign does ## \frac{\partial A\left(T\right)}{\partial T} ## have. My next task is to calculate ## c_V ## as a function of ## T ## and ## V ## and determine when it is volume independent.

    $$ c_V = \left(\frac{\partial U}{\partial T}\right)_V = - \frac{1}{V} N k_B \left(2 T \frac{\partial A\left(T\right)}{\partial T} + T^2 \frac{\partial^2 A\left(T\right)}{\partial^2 T}\right) $$

    If I look at this equation I cannot say when it does not depend on ## V ##. Where have I made a mistake?

    Any help would be greatly appreciated
     
  2. jcsd
  3. May 25, 2017 #2

    kuruman

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    Homework Helper
    Gold Member

    Your lower limit of integration should not be zero. The gas is expanding, but not from zero initial volume. When you integrate, you get ##-\frac{1}{V}##. When the gas expands, from initial volume ##V_i## to final volume ##V##, ##V_i~<V## and ##- \left[ \frac{1}{V} \right]_{V_i}^{V} >0##.
     
  4. May 25, 2017 #3
    Thank you, kuruman. I should have noticed that I can't use zero as the lower limit. If I use ## V_i ## then I get
    $$ U = - \left[\frac{1}{V}-\frac{1}{V_i}\right] N k_B T^2 \frac{\partial A}{\partial T} > 0 $$
    How do I determine now ## A ##, so that my ## c_V ## does not depend on ## V ##
     
  5. May 25, 2017 #4
    Your limits of integration should go from the current volume to infinite volume (where the ideal gas law applies). At the infinite volume limit, ##U=U^{IG}(T)##
     
  6. May 25, 2017 #5
    The internal energy is a total differential. In case of an ideal gas it depends only on the specific heat capacity by constant volume. In this case I calculate using the hint something like specific heat capacity by constant temperature. I mean ##c_T## in ## dU=c_V dT + c_T dV##. But I have no information about my first term. I must calculate the heat capacity by constant volume in the next part of the task.
     
    Last edited: May 25, 2017
  7. May 25, 2017 #6
    The internal energy in your problem is $$U(T,V)=C_vT-\frac{N k_b T^2}{V} \frac{d A}{dT}$$
     
  8. May 25, 2017 #7
    Thank you, Chestermiller. Maybe you think $$
    U(T,V)=C_vT +\frac{N k_b T^2}{V} \frac{d A}{dT} $$ since ## \int_V^{\infty} 1 / V^2 = - \left[1 / V\right]_V^{\infty} = - 0 + 1 / V ##
     
  9. May 25, 2017 #8
    No. Differentiate it with respect to V and see what you get.
     
  10. May 25, 2017 #9
    It is strange: $$ \frac{d\left( 1/V\right)}{dV} = -1 / V^2,$$ but it is also true, that $$
    \int_V^{\infty} 1 / V^2 = - \left[1 / V\right]_V^{\infty} = - 0 + 1 / V
    $$
     
  11. May 25, 2017 #10
    Flip the limits of integration.
     
  12. May 25, 2017 #11
    Thank you! To the second part of the task. I can define new ##C_V\left(T, V\right)##, so that if I let ##V## to be infinite (ideal gas), I will get the internal energy of the ideal gas. So it is just
    $$ C_V\left(V, T\right) = 3/2 N k_B - N k_B / V \frac{d}{dt}\left(T^2 \frac{d}{dt} A\right) $$
    It means, when ## dA/dT ## or ##
    \left(2 T \frac{\partial A\left(T\right)}{\partial T} + T^2 \frac{\partial^2 A\left(T\right)}{\partial^2 T}\right)
    ## are zero, I have no volume dependency.
     
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