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Combining gaseous spaces with PV= mRT

  1. Nov 23, 2011 #1
    I'm attempting to construct a very simplified mathmatical model of a stirling engine. I'm assuming that in one space hot gas exists at a temperature of 800k and in another the gas exists at its cold temperature of 300K.

    The gas is shuttled back from each space via displacer, which does so by changing the volume of each space... the spaces aren't sealed from each other (but i'm still assuming a strict temperature boundary between them)

    I want to calculate the total pressure of the system with air, forgetting for the moment about taking power out of the system with a piston..

    I suppose my first question is, with PV=mRT, when calculating m, do I take into account that the air density is different at the different temperatures, or is this accounted for already in the formula?

    And the second, when I have two pressures for each area, can I simply just combine the two to find the resultant pressure?

    Thanks :smile:
     
  2. jcsd
  3. Nov 24, 2011 #2
    Where did you get your equation? I am familiar with the classical equation of state for an ideal gas under conditions of equilibrium: PV=nRT

    Here P is the gas pressure in Pascals, V is the total gas volume in cubic meters, n is the total number of moles of the gas, R is the universal gas constant in Joules, and T is the temperature in Kelvins.

    Personally, I would convert this to P = nkT, where n is the molecular number density in molecules per cubic meter and k is Boltzmann's Constant. The number density, n, is easily calculated from: n = P/kT.

    How you would get the gases in your heat engine to remain under conditions of equilibrium is a real puzzler.
     
  4. Nov 24, 2011 #3
    PV=MRT is an equivalent form of the ideal gas law, except that R is the gas constant for the gas in question (e.g. air) and is not "universal" like the molar gas constant. The relation between the mass-specific gas constant and the universal molar gas constant is R = Runiv/w, where w is the molecular weight of the gas. For air, the gas constant is 53.35 ft-lbf/lbm R.

    Personally, my favorite form of the ideal gas law is P/ρ=RT....

    BBB
     
  5. Nov 24, 2011 #4
    The mass of air in volume V with pressure P and temperature T is m = PV/RT. The air density is ρ = m/V = P/RT. What was your question?

    Well, in general, in gas dynamics, you almost never are able to use the arithmetic mean of two pressures to find the result of a mixing process -- or almost any kind of process. (An exception would be heat transfer between two chambers of equal volume, with the same mass in each volume, where one chamber is initially warmer than the other.)

    Now, in this case, you say you "have two pressures for each area" -- and I have no idea what you mean here. Maybe a diagram, or at least a precise description of the sequence of states in each chamber, would help.

    BBB
     
  6. Nov 24, 2011 #5
    Sorry for my ambiguity, I guess I didn't explain it very well.

    Imagine a sealed cylinder, with a total volume of 1 m^3. The bottom half of the cylinder is being kept at a temperature of 300k. The top half is kept at a temperature of 900k.

    Now, there is a circular disc (of negligable volume, which does not conduct heat) suspended half way up in the cylinder, so that there is 1mm gap between it and the wall of the cylinder around the circumference (creating an annulus). We assume that because of the very small gap that there is no heat conduction between gasses, but the volume is still one and not separated.

    Does it follow that for this situation with the 0.5m3 to 0.5m3 volume split the pressure inside such a cylinder would be calculated using the mean of the two temperatures, i.e 600K.

    Using the same logic if it were a 0.25m^3 and 0.75m^3 volume split between hot and cold, would it follow that a weighted average could be used? I.e 0.25*900 + 0.75*300 = average temp of 450K?

    I know saying that no heat transfer between the two is a big assumption, but for my model it is nessesary at this stage, but is the way I'm calculating the average gas temperature (which i'm then using to find the pressure using PV=mRT) correct?
     
  7. Nov 24, 2011 #6
    You haven't specified enough information to be able to calculate the pressure. You need to know how much gas is in the chamber. Suppose we know the total mass is M, split between chamber 1 and chamber 2 so that

    M = m1+m2

    We will stipulate that the pressure is the same in chamber 1 and chamber 2:

    P = P1 = P2

    Solving the ideal gas law for the mass of gas:

    m = PV/RT

    Then the same relation gives

    m1 + m2 = PV1/RT1 + PV2/RT2 = M

    Thus the pressure is

    [itex]P = \frac{MR}{V_{1}/T_{1} + V_{2}/T_{2}}[/itex]

    If you want the mass in chamber 1, for example, then take this result and multiply by V/RT:

    [itex]m_{1} = \frac{MV_{1}/T_{1}}{V_{1}/T_{1} + V_{2}/T_{2}}[/itex]

    So the answer is no, you don't interpolate the temperatures using the volume as a weighting factor. You actually interpolate the inverse temperature 1/T. My advice is to do the algebra. It's not hard.

    BBB
     
  8. Nov 25, 2011 #7
    Forget R, V, and m. None of these parameters are necessary to obtain the pressure on the partition (which, of course, will be the same throughout the chamber). Simply use the modern equation of state: P = nkT. Since the temperature in the upper chamber is three times the temperature in the lower chamber, the number of molecules in the upper chamber must be one-third the number in the lower chamber. If you know the total mass or the mean molecular mass, the rest is simple algebra.

    The mean temperature of the two chambers is obtained in basically the same way: three parts 900° and one part 300° = 750°.

    See my post#2 for parameter definitions. It would be a big help if all posters would define their terms in each of their equations and give the units of measurement.
     
  9. Nov 26, 2011 #8
    Ok, bbbeard, so using your equation:

    [itex]P = \frac{MR}{V_{1}/T_{1} + V_{2}/T_{2}}[/itex]

    With:

    V1 = 0.75 m^3
    V2 = 0.25 m^3
    T1 = 300 k
    T2 = 900 k
    M = 1kg
    R = 287 J/Kgk

    Which gives an answer for P of 103,320 Pa

    Whereas following klimatos logic, my T would be 0.75*300 + 0.25*900 = 450

    Thus using PV = mRT, gives P= (1*287*450)/1 =129,150


    So, who is right? :D
     
  10. Nov 26, 2011 #9
    You wrote: "Forget R, V, and m. None of these parameters are necessary to obtain the pressure on the partition...."

    Please explain how you find the pressure without knowing the amount of gas in the system and the gas constant (note the Boltzmann constant k is just the gas constant per particle).

    And why do you call P=nkT the "modern" equation of state? This is just another version of the ideal gas law. [Kalus: n is the particle density = the number of gas molecules per unit volume]. This is entirely equivalent to the other forms that have been posted in this thread.

    Maybe you want to go over that again. You wrote "Since the temperature in the upper chamber is three times the temperature in the lower chamber, the number of molecules in the upper chamber must be one-third the number in the lower chamber. " So why are you asserting

    Tmean = (3TU+1TL)/4 = 750

    instead of

    Tmean = ((1/3)TU+1TL)/(4/3) = 450

    ?

    BBB
     
    Last edited: Nov 26, 2011
  11. Nov 26, 2011 #10
    I don't think you understood what Klimatos wrote. First, what he proposed was a correct line of thinking that concluded that the amount of gas in the 900 K chamber was 1/3 the amount in the 300 K chamber, for the case where the volumes are equal: V1=V2=0.5 m3. If the volumes are not equal, then you get a different ratio e.g. if the hot chamber is 1/3 the volume of the cold chamber, then the hot chamber has 1/9 the amount of gas as the cold chamber. If the volumes are different from each other, it's best to use a form of the ideal gas law that contains the volume explictly:

    e.g. instead of using P = nkT, use PV = NkT
    (n = particle number density, N=number of particles, and N/V=n)

    or

    e.g. instead of using P = ρRT, use PV = MRT.
    (ρ = mass density, M = mass of gas, and M/V=ρ)

    So for

    V1 = 0.75 m^3
    V2 = 0.25 m^3
    T1 = 300 k
    T2 = 900 k

    we know

    [itex]\frac{M_1}{M_2}=\frac{V_1}{V_2}\frac{T_2}{T_1}[/itex]

    [itex]\frac{M_1}{M_2}=9[/itex]

    so the mean temperature is

    Tmean = (9TL+1TU)/10 = 360.


    Then using PV = MRT, this gives P= (1*287*360)/1 =103,320 -- which is the same result that I derived. Which method seems simpler to you? (It doesn't really matter, because they are equivalent for this problem.)

    BBB
     
    Last edited: Nov 26, 2011
  12. Nov 26, 2011 #11
    You are correct in that the mean gas temperature should be 450K. I apologize for trying to post while coming down with a head cold. Please give me a day to clear my brain and allow me to try again.
     
  13. Nov 26, 2011 #12
    The following sets of parameters appear to satisfy both the postulates (except for the volume split) and the gas laws:

    Total cylinder volume is one cubic meter and total gas mass is one kilogram.

    Upper Cylinder: P = 129,150 Pascals, V = 0.5 m3, T = 900K,
    M = 0.25 kg, N = 5.20 x 1024 molecules, n = 1.04 x 1025 molecules per cubic meter.

    Lower Cylinder: P = 129,150 Pascals, V = 0.5 m3, T = 300K,
    M = 0.75 kg, N = 1.56 x 1025 molecules, n = 3.12 x 1025 molecules per cubic meter.

    Mean gas temperature is 450K

    PV = NkT, P = nkT, PU = PL, NU + NL = 2.08 x 1025 molecules.

    Comments and queries please.
     
  14. Nov 27, 2011 #13
    Thank you both for your replies :smile:

    I've worked it through algebraically with both your methods and I see how it works now :)

    I should've started the problem algebraically in the first place, instead of jumping in there with the numbers.

    It's really helped with my project, so thank you :)
     
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