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Thermodynamics: air expansion in a cylinder

  1. Mar 27, 2016 #1
    1. The problem statement, all variables and given/known data
    Consider air expansion inside a cylinder. Assume that the volume and initial pressure is 1 ft^3 and 1500 PSI ABS respectively. If the expansion process is reversible and the path is given by P.V^1.4=constant. Calculate the total work done by the gas to reach the final volume of 8 ft^3, express the result in BTU if used:
    a) The Van der Waals equation

    2. Relevant equations
    (P + a^2 / v^2) * (v - b) = RT

    (V−b)T^R/CV=constant.

    3. The attempt at a solution
    T1(v1 − b)^ R/cV = T2(v2 − b)^ R/cV
    W = −∆E = cV (T2 − T1) − a(1/v2 − 1/v1)
    how do I replace T2 and T1?
    I have two ideas: T1= (P + a^2 / v^2) * (v - b) /R
    T2 = (P + a^2 / v^2) * (v - b) /R
    or
    T1= (P + a^2 / v^2) * (v - b) /R
    T2= T1 ( (v1 − b)^ R/cV / (v2 − b)^ R/cV )
     
  2. jcsd
  3. Mar 27, 2016 #2
    If you know the P-V relationship for the process, why do you need the van der Waals equation to calculate the work?
     
  4. Mar 27, 2016 #3
    I think that the process is for ideal gas, and i already did that, but for van der waals it couldn't be the process, how can I find the work without that process?
     
  5. Mar 27, 2016 #4
    You don't know anything about the heat added or removed, so you can't do a van der waals gas. If you assume that the process is adiabatic and reversible, then you can. But why do you think that the problem statement is referring to an ideal gas, particularly if the initial pressure is 100 atm?
     
  6. Mar 27, 2016 #5
    yes, I have to assume that the process is adiabatic and reversible, and because the first point was find the work with an ideal gas
     
  7. Mar 27, 2016 #6
    Did you leave that out of the problem statement? What else did you leave out?
     
  8. Mar 27, 2016 #7
    the problem only says to find the work with an ideal gas and with the van der waals equation, but I already did with an ideal gas, i don`t how to do it with the van der waals equation
     
  9. Mar 28, 2016 #8
    The starting point for the analysis is the general equation for dU as a function of dT and dV:
    $$dU=C_vdT+\left[T\left(\frac{\partial P}{\partial T}\right)_V-P\right]dV$$
    If you substitute the equation for a van der waals gas into the second term in this equation, what do you get for dU?
     
  10. Mar 28, 2016 #9
    U = nCv (T2- T1) + a (1/ v1 - 1/v2)
     
  11. Mar 28, 2016 #10
    U= nCv (T2-T1) + a (1/v1 - 1/v2)
     
  12. Mar 28, 2016 #11
    I was asking for dU in terms of dT and dV. When you integrated, how did you know that Cv was not a function of V? From your equation for dU in terms of dT and dV, can you tell if Cv is a function of V? How?
     
  13. Mar 28, 2016 #12
    A trick on this problem is to recognize that , even though you don't know the number of moles n or the initial temperature T1, you know the product nT1. This will help you get the amount of work by determining delta U.
     
  14. Mar 28, 2016 #13
    So how do I know if Cv is a function of V?
     
  15. Mar 29, 2016 #14
    You found that, for a van der waals gas, $$du=c_vdT+\frac{a}{v^2}dv$$where I have used lower case symbols to represent the properties per mole. From this equation,$$\left(\frac{\partial u}{\partial T}\right)_v=c_v$$and $$\left(\frac{\partial u}{\partial v}\right)_T=\frac{a}{v^2}$$So that means that $$\frac{\partial^2 u}{\partial T \partial v}=\left(\frac{\partial c_v}{\partial v}\right)_T=\left(\frac{\partial (a/v^2)}{\partial T}\right)_v$$But, the right hand side of this equation is equal to zero. Therefore, ##c_v## is not a function of v for a van der Waals gas; it is only a function of T. Therefore, it must be equal to the limiting value at infinite v, which denotes the ideal gas value. Therefore ##c_v## for a van der Waals gas is the value for the same gas in the ideal gas limit.

    Sara, I think I may have been incorrect in what I said in post #12. For a van der Waals gas, it may not be possible to get the total work done W, but only the work per mole w (unless they told you the number of moles or the initial temperature). So let's work it out per mole first (anyway).

    You found that $$w=-\Delta u=-\left[c_v(T_2-T_1)+a\left(\frac{1}{v_1}-\frac{1}{v_2}\right)\right]$$
    You should also be able to show that $$\frac{T_2}{T_1}=\left(\frac{(v_1-b)}{(v_2-b)}\right)^{\frac{R}{c_v}}$$
    Therefore, $$w=-\left[c_vT_1\left(\left(\frac{(v_1-b)}{(v_2-b)}\right)^{\frac{R}{c_v}}-1\right)+a\left(\frac{1}{v_1}-\frac{1}{v_2}\right)\right]$$
    This is the work per mole. But, I don't know how to get the number of moles without knowing T1.

    Chet
     
  16. Mar 29, 2016 #15
    To replace T in the equation like
    T=(P + a^2 / v^2) * (v - b) / R
    And put everything in terms of P and V
     
  17. Mar 29, 2016 #16
    The v's in this equation are the volume per mole, not the actual volume. So, vn = V. But you don't know n.

    Also, in your equation, there should be an a, not an a2.
     
  18. Mar 29, 2016 #17
    Yes, it was my mistake
     
  19. Mar 29, 2016 #18
    But I can take n=1
     
  20. Mar 29, 2016 #19
    Did they tell you that there is 1 mole?
     
  21. Mar 29, 2016 #20
    No, but I can assume that
     
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