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Thermodynamics: Finding the molar heat capacity

  1. Jun 18, 2013 #1
    1. The problem statement, all variables and given/known data
    An ideal gas has a molar heat capacity ##C_V## at constant volume. Find the molar heat capacity of this gas as a function of its volume ##V##, if the gas undergoes the following process: ##T=T_0e^{\alpha V}## here ##T_0## and ##\alpha## are constants.


    2. Relevant equations



    3. The attempt at a solution
    Work done by gas: ##\displaystyle W=\int PdV##
    From ideal gas equation: ##PV=nRT=nRT_0e^{\alpha V}##. Hence
    [tex]W=\int_{V_1}^{V_2} \frac{nRT_0 e^{\alpha V}}{V}dV[/tex]
    But I cannot integrate this. :confused:
     
    Last edited: Jun 18, 2013
  2. jcsd
  3. Jun 18, 2013 #2
    no work is done if the volume is kept constant - sorry I jumped too quickly on this.
     
  4. Jun 18, 2013 #3
    Yes but the question does not state that the volume is constant during the process.
     
  5. Jun 18, 2013 #4
    It might help to integrate by parts.
     
  6. Jun 18, 2013 #5
    I plugged this in wolfram alpha: http://www.wolframalpha.com/input/?i=integrate+e^(ax)dx/x

    I don't think this question requires me to use Exponential Integral as shown by wolfram alpha.
     
  7. Jun 18, 2013 #6

    ehild

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    Ask yourself: What is heat capacity?

    ehild
     
  8. Jun 18, 2013 #7
    Heat required to change the temperature of a body by a given amount.

    I approached the problem the same way as I have done for the polytropic processes. I calculate the work done in those processes through integration in terms of ##\Delta T##, change in temperature. I then write the change in internal energy in terms of ##\Delta T## and plug them in ##Q=W+\Delta U##. Rewriting Q in terms of heat capacity, I obtain the answer. Is it incorrect to use the same approach here?
     
  9. Jun 18, 2013 #8

    ehild

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    Change by what amount? Can you formulate the definition of heat capacity mathematically?

    How do you write Q (the total amount of heat during a process) in terms of heat capacity, when it changes during the process? Q depends on V and you integrate with respect to V.

    Have you solved the problem? What is your solution?

    ehild
     
  10. Jun 18, 2013 #9
    Sorry ehild, but I am less familiar with the technical terms. :redface:

    [tex]C=\frac{\Delta Q}{n\Delta T}[/tex]

    Is it wrong to write ##Q=nC\Delta T## where C is the molar heat capacity?
     
  11. Jun 18, 2013 #10

    ehild

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    C can depend on other state variables. Here it depends on V, and also on T as T is function of V. The total heat Q taken in a process is not necessarily proportional to ΔT. The heat taken by the system depends also on the process. How do your lecture notes or books define the molar heat capacity? Certainly not as you said that "Heat required to change the temperature of a body by a given amount." If the given amount is 218 K, and you need 1000 J heat to change the temperature by that amount than the heat capacity is 1000 J????


    ehild
     
  12. Jun 18, 2013 #11
    Sorry, I just looked up heat capacity in my book. It defines heat capacity as C=Q/ΔT.

    I am still not sure how to begin with the problem. :confused:
     
  13. Jun 18, 2013 #12

    ehild

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    The same way as you define speed as ds/dt, you can define heat capacity at a given state as small amount of heat divided by small change of temperature C=δQ/dt . The First Law of Thermodynamics says that dU=δQ+δW, that is, δQ=dU-δW. The small amount of heat taken is equal to the sum of the infinitesimal change of internal energy + infinitesimal amount of work.

    dU=CvdT, δW=-pdV - write this in terms of V and dT.


    ehild
     
  14. Jun 18, 2013 #13
    I can replace P with ##nRT/V=nRT_0e^{\alpha V}/V## and ##dV## with ##dT/(T_0 \alpha e^{\alpha V}##

    $$dW=\frac{nRdT}{\alpha V}$$
     
  15. Jun 18, 2013 #14

    ehild

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    Very good, plug into the equation δQ=dU-δW and find C=δQ/dt. (Recall that it is an ideal gas, so dU=Cvdt.)

    ehild
     
  16. Jun 18, 2013 #15
    [tex]CdT=C_VdT+\frac{nRdT}{\alpha V}[/tex]
    I can cancel dT but what about n? :confused:
     
  17. Jun 18, 2013 #16

    ehild

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    The molar heat capacity is the heat capacity of one mole gas. n=1.
     
  18. Jun 18, 2013 #17
    Completely missed that, thanks! :smile:
     
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