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Homework Help: Work in thermodynamic processes

  1. Apr 6, 2010 #1
    1. The problem statement, all variables and given/known data

    Two moles of an ideal monoatomic gas trebles its initial volume in an isobaric expansion from state A to state B. The gas is then cooled isochorically to state C and finally compressed isothermally until it returns to state A. The molar gas constant is R = 8.314 J mol–1 K–1 and Boltzmann's constant is 1.38 × 10–23 J K–1.

    State B corresponds to a pressure P = 8 atm (1 atm = 1.013 × 105 Pa) and temperature T = 552°C. (I also know that temperature of the gas in state A is 275 K)

    What is the work done on the surroundings in each of the following processes:

    (i) A→B

    (ii) C→A

    3. The attempt at a solution

    (i) This process is isobaric
    p = constant = pB


    I could calculate the volumes from the given information. But it easier to substitute pV = nRT for each state and get:
    [tex]W = nR(T_B - T_A)[/tex]
    W = 8(1.013 × 105) ∙ (8.314).(825.15 K - 275.05 K)
    W= 3706389847 J

    Why is the value for work so large? The correct answer to this part must be 9150.

    Since we have an isothermal process:
    [tex]pV = nRT = constant = nRT_C[/tex]

    W = nRTA ∙ln(VA/VC)

    W = 8(1.013 × 105) ∙ 8.314 ∙ 275K ∙ ln (VA/VC)

    I'm stuck. How do I get the volumes for states A and C? :confused:

    (P.S. I also know that the work done in the processe B→C is 0, because it is an isochoric (i.e. isovolumetric) process).
  2. jcsd
  3. Apr 6, 2010 #2
    You can find VA directly from the given information, as you yourself stated in (i).

    Next is state C. The gas is cooled isochorically from B to C. What does that tell you?
  4. Apr 6, 2010 #3
    It tells me that the volume of state C is equal to the volume of state B. So I just need to find the volume for A and B(=C).

    So for the volume of C:


    [tex]\frac{8(1.013 \times 10^5)V}{825 K}=n(8.314)[/tex]

    What value do I need to use for the number of moles?
  5. Apr 6, 2010 #4
    Look through the problem statement. :wink:
  6. Apr 6, 2010 #5
    I already have. If I use 8 (1.013 × 105) for the pressure, I can't use it again for the number of moles!
  7. Apr 6, 2010 #6
    What do the two first words say?
  8. Apr 7, 2010 #7
    Wow, that was dumb of me. I'm trying to do a lot of things at the same time :shy:
    I managed to get part (i) right using the right value for n.

    But for part (ii):

    8(1.013 × 105)V=2(8.314)(825 K)


    Volume of C is 0.016. And the problem says "gas trebles its initial volume in an isobaric expansion from state A to state B", so that means the volume at C is 3 times larger than that of A:

    [tex]V_A=\frac{V_C}{3} = \frac{0.016}{3}=5.3 \times 10^{-3}[/tex]

    Are these values correct?

    W = 2 ∙ 8.314 ∙ 275K ∙ ln (5.3 \times 10^{-3}/0.016)


    does this answer look alright? :rolleyes:
  9. Apr 7, 2010 #8
    I haven't done the numeric calculation myself, but your method checks out.

    Efficiency tip: Since you know the gas triples in volume, you don't really need to calculate the volume of before/after for the isothermal compression. You know that

    [tex]w=nRT\ln\left(\frac{V_A}{V_C}\right)=nRT\ln\left(\frac{V_A}{3V_A}\right)=-nRT\ln 3[/tex]

  10. Apr 8, 2010 #9
    Thanks a lot for the tips! :smile:

    This is my last question: The gas in state A now expands adiabatically to 7 times its original volume (State D). What is the final temperature of the gas?

    I tried to use the equation of state for an ideal gas PV=nRT to find the temprature:



    I know that in an adiabatic process Q=0 meaning that no energy enters or leaves the system by heat. We don't know whether or not the pressure is constant, but since I don't know the final pressure I assume it is constant


    When I substitute the values and solve for the Tf my answer is wrong. Am I using the wrong equation?
  11. Apr 8, 2010 #10
    For adiabatic processes, you need the adiabat equation

    [tex]T_1V_1^{\alpha}=T_2V_2^{\alpha}[/tex] where [tex]\alpha[/tex] is the number of degrees of freedom divided by 2, which is [tex]\alpha=\frac{3}{2}[/tex] for a monoatomic ideal gas, or [tex]\alpha=\frac{5}{2}[/tex] for a diatomic ideal gas.

    Can you take it from there?
  12. Apr 8, 2010 #11
    Thank you very much, that was very helpful. I really appreciate your help.
  13. Apr 9, 2010 #12
    No problem. :-)
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