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Help with Thermodynamics

  1. Jul 22, 2004 #1
    figure for problem

    I am trying to figure out this problem and could use a little bit of help. Escuse me if my LaTex is bad, this is my first time.

    One mole of an ideal monatomic gas is caused to go through the cycle shown in the figure (all processes are reversible). Express all answers in terms of the pressure [tex] p_0 [/tex] and the volume [tex] V_0 [/tex] at a point [tex] a [/tex] in the diagram.

    a) How much work is done on the gas in expanding the gas from a to c along the path abc?

    We can work out this by summing the work from ab and bc. ab is an isobaric process, so W = [tex] P ( V_2 - V_1 ) [/tex] ---> [tex] P_0 ( 4 V_0 - V_0 ) [/tex] = [tex] 3 V_o p_o [/tex]

    bc is an isochoric process, so no work is done. The final answer is as above.

    That one wasn't so bad.

    b) What is the change in internal energy and entropy in going from b to c?

    This is isochoric process. [tex] \Delta U = m C_v \Delta T [/tex] and [tex] \Delta S = \int \frac {dQ} {T} [/tex]

    My problem is that I don't really understand where to go from here. since [tex] C_v [/tex] is throwing me off a bit. How to solve these in terms of what is asked for? And with entropy, can I find a ratio of T in terms of pressure and volume? What do I really need to do here?

    c) What is the change in internal energy and entropy in going through one complete cycle?

    Since this is a cyclic reversible process that starts and ends at a, the change in internal energy and the entropy must both be 0. Do I need to say more? Should I prove this or is it just too obvious?

    Thanks in advance for your help, and any comments are welcomed :-)
     
    Last edited: Jul 22, 2004
  2. jcsd
  3. Jul 22, 2004 #2
    Maybe I am completely wrong about this. If someone has knowledge of part b, please let me hear your input...
     
  4. Jul 22, 2004 #3

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    Is it fair to say that for an isochoric process b--> c, the ideal gas law gives you that Delta T is proportional to V Delta p where in this case V is 4 V_o and Delta p is p_o ? Then, with the right proportionality constant (which I leave to you), you can substitute that value of Delta T into your Delta U = m C_v Delta T.

    Note: My dog is bigger than your dog, and my LaTex is worse than your LaTex. I just spent 10 minutes trying to work out the kinks in the above, and gave up and just presented it in ugly non-LaTex form.
     
  5. Jul 23, 2004 #4
    For a monotonic ideal gas Cv = 3/2R. Use the the ideal gas eqation to change from temperature to pressure and volume. Since in part b you have an isochoric process, the volume will be fixed and only the pressure will change. This will give you the following equation

    [tex]\Delta U = 1.5m\Delta{p}V = 1.5mp_0V[/tex]

    Note: The problem told us that there is one mole of gas, so the n doesn't appear in the equation. I know most of the mathematical details were skipped, but if you need to see them let me know.


    To find the entropy we should first start from the following general entropy equation

    [tex]dS = \frac{dU}{T} + \frac{P}{T}dV = \frac{C_V}{T}dT + \frac{R}{V}dV[/tex]

    I made use of dU = CvdT and P = RT/V

    Integrating the expression gives

    [tex]\Delta S = C_V ln\frac{T_2}{T_1} + Rln\frac{V_2}{V_1}[/tex]

    Substituting the ideal gas relation for T and neglecting the second term for an isochoric process gives

    [tex]\Delta S = C_Vln\frac{P_2}{P_1}[/tex]

    Using Cv = 3/2R and P2 = 2P1

    [tex]\Delta S = \frac{3}{2}Rln2[/tex]

    By the way, this derivation for entropy was done the rigorous way and only simplified later. Using this method, you can calculate the entropy change from c to a. If you're curious, the expression for this case would be

    [tex]\Delta S = (C_V+R)ln\frac{V_2}{V_1}+C_Vln\frac{P_2}{P_1} = 2.5Rln\frac{1}{4}+1.5Rln\frac{1}{2}[/tex]

    In part c, you might want to show how it works out mathematically.
     
  6. Jul 23, 2004 #5
    Since it is monatomic ideal gas, we know that [tex] C_v = \frac {3 R} {2} [/tex], where R is the universal gas constant. But now things are in terms of R and m still exists out in front! Can we relate m and R in terms of V or p?
     
  7. Jul 23, 2004 #6
    I was typing while someone else posted!

    Thanks for input! That really makes much more sense now...
     
  8. Jul 23, 2004 #7
    We could convert the m by saying it is equivalent to moles*(molecular weight) and since n =1, then m = MW. It shouldn't be a problem if your answer has MW and R because they are both constants.
     
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