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Pressure and force in a cylinder

  1. May 7, 2015 #1
    A cylinder containing oxygen gas O2 has a piston that can move unfazed by friction. The piston has a diameter of d = 16cm. The oxygen gas has a temp of 40,0 degrees Celsius and a pressure of 145kPa. The pressure outside the cylinder is 101kPa. The piston is held in a stationary position s = 60,0cm from the end wall of the cylinder by applying a constant force F that stands perpendicular on the piston.

    1) How big must F be to keep the piston in a constant position?
    2) Calculate the amount of molecules in the gas.
    3) Assume that F gradually reduces to zero, so that the volume of the gas expands until it reaches the same value as the pressure on the outside. The temperature of the gas is held constant. How big is the distance s from the end wall now?
    4) The cylinder is held over a flame until the temperature of the oxygen gas has raised to 60 degrees Celsius. We let the gas expand so the pressure inside the cylinder remains constant. Use the first law of thermophysics to calculate the change in the internal energy of the gas.

    Specific heat capacity of oxygen gas is c = 923J / kg*K

    Okay so...

    The work performed by a gas on the surroundings during volume change at a constant pressure is given by:
    W = -pΔV

    p = pressure

    ΔV = Volume change in the gas

    Any ideas? :oldconfused:https://s.yimg.com/hd/answers/i/e5c1cf23c7a0446294a90a995aea323c_A.png?a=answers&mr=0&x=1431033458&s=a914c422328287076bc156ca5e12a6a0 [Broken]
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. May 7, 2015 #2
    Yes. I definitely have an idea. Why don't you start out by showing us how you do the easy parts (1), (2), and (3) before you start working on part (4)?

    By the way, is that the heat capacity value at constant pressure or the heat capacity value at constant volume?

    Chet
     
  4. May 7, 2015 #3
    a)

    p = F/A

    F = pA = 145kPa(60*16)cm = 139200N

    b)

    pV = NkT

    V = pi*r^2*h = 48230 cm^3

    N = pV/kT => N = (145kPa*48230cm^3) / (1.38*10^-23J/K * 40) = 1.266*10^28

    I think those are right but I'm confused by the last two.
     
  5. May 7, 2015 #4
    Both answers are incorrect. Do a force balance on the piston. What are the forces (other than F) acting on the faces of the piston. What is the area of the piston? (You calculated the area A incorrectly when you determined the initial volume V). In applying the ideal gas law, shouldn't the volume be expressed in cubic meters?

    Chet
     
  6. May 8, 2015 #5
    Yeah, and I also think the degrees must be in Kelvin, which gives 273 + 40 = 313K

    If the volume must be in m^3 we get 0.04823 m^3, but I don't understand what you mean by A being incorrect. It's a cylindrical piston, so the area of a circle is pi*r^2, and times the height h = 60 gives us the volume, right? Volume of a cylinder is pi*r^2*h = A*h.

    So with those numbers I get:

    N = (145kPa * 0.04823 m^3) / (1.38*10^-23J/K * 313K) = 0.01619 * 10^23

    ??

    Wrong area on the first one so:

    F = pA = 145000 N/m^2 * ((3.14*16^2)cm^2)*10^-4 m^2 = 11655.7 N
     
    Last edited: May 8, 2015
  7. May 8, 2015 #6
    The diameter of the cross section is 16 cm. the radius is only 8 cm.

    Chet
     
  8. May 8, 2015 #7
    Aaaah, simple mistake, my bad! :approve:
     
  9. May 8, 2015 #8
    OK. What are your thoughts on part 3?

    Incidentally, the force F that you need to apply to the piston to hold it in equilibrium should be F = (145000-101000)A, not 145000A

    Chet
     
  10. May 9, 2015 #9
    I'll use the equation of state to find the new volume:

    (p1v1)/T1 = (p2V2)/T2

    Temperature remains the same in this instance so T1 = T2

    Which leaves us with: p1V1 = p2V2

    I wish to find the new volume V2, so V2 = (p1V1)/p2 = ((145kPa * (pi*(0.080m)^2*0.600m))/101 kPa = 0.0174m^3

    Now using this information to find the new length s:

    V2 = pi*r^2*s => s = V2/pi*r^2 = 0.0174m^3/pi*(0.080m)^2 = 0.86m
     
  11. May 9, 2015 #10
    With regard to part 4, what is the effect of temperature on the internal energy of an ideal gas? At a given temperature, Is the internal energy of an ideal gas affected by its pressure or specific volume?

    Chet
     
  12. May 9, 2015 #11
    Right. So the sum of forces must be: pA - F - p0A = 0

    To keep the piston in equilibrium, the forces of F + air pressure must equalize the pressure from the oxygen gas.

    F = pA - p0A = (p-p0)A

    p0 = 101kPA, area of the piston is equal to that of a circle: A = pi*r^2

    Therefore: F = (145-101)kPa * pi * (0.080m)^2 = 0.88 kN
     
  13. May 9, 2015 #12
    I didn't check the arithmetic, but this is the correct approach.

    Chet
     
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