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Ideal gas law + Newton's 2nd law problem

  1. Jan 27, 2007 #1
    1. The problem statement, all variables and given/known data
    A cylinder has its lid connected to a spring with k = 2000 N/m. The cylinder is filled with 5.00 L of gas witht he spring relaxed at a pressure of 1.00 atm and a temperature of 20.0C. (a) If the lid has a cross-sectional area of 0.0100 m2 and negligible mass, how high will the lid rise when the temperature is raised to 250C? (b) What is the pressure of the gas at 250C. Diagram of this picture is attached

    2. Relevant equations
    PV = nRT
    F = PA
    F = -kx


    3. The attempt at a solution
    In looking that the diagram, I identified three forces experted on the spring. The normal force by the air, spring force and weight force. Since mass is neglitable, there is only the spring and normal force. Therefore, my equation came out as:
    N = S
    PA = kx
    Just by looking at the question, I know we have A and k given in the question, and what we are looking to solve is x. Therefore, in an attempt to get P, I used the ideal gas law:
    P1V1 / T1 = P2V2 / T2
    1atm * 5.00L / 293K = P2V2 / 523K
    P2V2 = 8.92J

    Now what do I do? In order to get pressure, I need volume, which is an unknown at this moment... Any clue or help would be useful, thanks!
     

    Attached Files:

    Last edited: Jan 28, 2007
  2. jcsd
  3. Jan 27, 2007 #2
    Maybe expressing delta V=P2*K*x and V2=V1+deltaV. Just a stab off the top of my head.
    JS
     
  4. Jan 27, 2007 #3
    V = P*k*x? Where did you get that equation from? I haven't seen an equation like that? Is it a derived equation? If so, where and how did you derive it?
     
  5. Jan 28, 2007 #4
    Sorry, was a bit hasty as I was cooking, the forces on the diaphragm at equilibrium:

    F=-KX is the equal and opposite to p2*A, hence Kx=P2*A or x=P2*A/k

    and delta V= A*x, does that help?
    JS
     
  6. Jan 28, 2007 #5

    Gib Z

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    I have no idea but to test the V=pkx, check the units and see if they match up :).

    If your friends ask, and you wanna sound smart, say you solved it by Dimensional Analysis :D
     
  7. Jan 28, 2007 #6

    Gib Z

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    Wait no, ignore me now>.<
     
  8. Jan 28, 2007 #7
    This is a tricky problem as you do not have V2 =). However, you can substitute near the end once you find out all the equations =).

    However, think logically. It takes 2000 N to push that block up. Hopefully this helps.
     
  9. Jan 28, 2007 #8
    Well I'm hopeful, as delV=A^2*p2/K which ends up in length ^3 units, so

    basically, by subbing P2(V1+delV)=nRT, and n can be solved from the initial conditions, so

    P2(v1+P2(a^2/k))=nRT v1=5L. area and k are known, then once P2 is known, go back and get x. I think this is right, but its late and we had a great bottle of chianti. :zzz: Likely there are more elegant soln's but hey this is my first day here after not thinking about physics for 20 years. But I'm already addicted to this site, would have killed for it during my student days.
    JS
     
    Last edited: Jan 28, 2007
  10. Jan 28, 2007 #9

    Andrew Mason

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    The pressure on the air in the cylinder is given by the spring and the atmospheric pressure. Initially, there is no spring force.

    [tex]P_iV_i = nRT_i[/tex]

    from that you can work out n.

    Now, as T increases and V increases, the spring is compressed which increases the pressure.

    [tex]P_fV_f = nRT_f[/tex]

    Since Vf = Vi + Ax and Pf = Pi + kx/A, this can be rewritten:

    [tex](P_i + kx/A)(V_i + Ax) = nRT_f[/tex]

    Try and work that out to get a solution for x.

    AM
     
    Last edited: Jan 28, 2007
  11. Jan 28, 2007 #10
    Hey, thanks pal. This really helped a lot. denverdoc, I completely understand your explanation. But Andrew, regarding your equation, I do not get how you derived this equation Pf = Pi + kx. If you can explain it to me, that would be great. Unit of pressure is in atm or N/m^2 and unit of kx is N or kg*m^2/s^2. These two units don't match up and therefore cannot be added
     
    Last edited: Jan 28, 2007
  12. Jan 28, 2007 #11

    Andrew Mason

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    Quite right. It should be Pi + kx/A. I have fixed it in my post above.

    AM
     
  13. Jan 28, 2007 #12
    Sorry, but I'm really bad at this. I am still confused about how you get the new equation Pf = Pi + kx/A. If Pf*A = kx, then wouldn't Pf = kx/A? Or are you saying I put down the wrong equation.
     
  14. Jan 28, 2007 #13
    Hey denverdoc, I used your method and I got 1.79atm for the pressure and 8.95*10^-6m for x. I think the pressure is right but x is wrong, since the value is too little to be noticeable in a real life experiment. This is what I did:
    N = S
    PA = kx
    Just by looking at the question, I know we have A and k given in the question, and what we are looking to solve is x. Therefore, in an attempt to get P, I used the ideal gas law:
    P1V1 / T1 = P2V2 / T2
    1atm * 5.00L / 293K = P2V2 / 523K
    P2V2 = 8.92J
    P2(v1+P2(a^2/k)) = 8.92
    P2(5.00 + P2 (0.01^2 / 2000)) = 8.92
    P2 = 1.79atm, -1*10^8atm (rejected)

    P2A = kx
    1.79 * 0.01^2 = 2000x
    x = 8.95*10^-6m

    I do not think there is an error in my arithmetics and algebra. I double checked them. But I may have made a mistake
     
    Last edited: Jan 28, 2007
  15. Jan 28, 2007 #14
    When I did your method, I got this:
    [tex](P_i + kx/A)(V_i + Ax) = nRT_f[/tex]
    (1 + 2000x/0.01) + (5 + 0.01x) = 8.93
    5 + 0.01x + 1000000x + 2000x^2 = 8.93
    2000x^2 + 1000000 -3.9 = 0
    x = (.1*10^6 +/- sqrt (1*10^12 + 31200)) / 4000
    x = -500m, 3.9*10-6m
    For this equation, I also got too small of a number. Is x supposed to be very small?
     
  16. Jan 28, 2007 #15
    Its a stiff spring, but 4 or 9 microns seems small indeed. what is the initial height of 5L at 20 degrees? .01 m^2 =100 cm^2 is about 2 inch radius for nearly gal and a half volume. I'm wondering about units--atmospheres need to be converted to N/m^2.
    JS
     
  17. Jan 28, 2007 #16
    P = 1.79 atm, x = 0.907m. Sounds much better. Thank you!!!:surprised
     
  18. Jan 28, 2007 #17
    No sweat, and truly thanks to perverse genes that thinks physics is recreational, my pleasure. Let us know if we got it right.
    Also be curious whether the two solns offered were in agreement.
    John
     
  19. Jan 28, 2007 #18
    That pressure doesn't seem right cy19861126 =).

    The pressure should be decreasing a tiny bit due to the new volume expansion. Simply, if you say the volume were to not change you'd get a 1.78 atm pressure =). (P1 * T2 / T1) = P2

    However, the spring is pushing on the gas and the gas is pushing on the spring.

    However, let's say hypothetically it went from 0.005 m^3 to 0.007 m^3.

    P1V1T2 / T1V2 = P2. P2 ~ 1.27 atm.

    Oh to get an idea of the height. 1.013 x 10^5 N/m^2 * 0.0100 m^2 = 1013 N. 1013 N / 2000 N (per meter) = 0.5065 meters is the height it's being pushed up initially. 0.5065 m * 0.0100 m^2 = 0.005065 m^3 ~ 5.065 Liters

    5 Liters = 0.005 m^3

    However, I guess we can look at the before and after picture to get an idea of delta h. Let's just say how much a spring gets pushed on when there's 1 atm. Then say the volume increased and find the new pressure. From there and then say how much that pressure pushes up on a spring.

    P1 / T1 = P2 / T2; P1 * T2 / T1 = P2.

    ( P2 * Area ) / k = x2. x2 - x1 = delta h.

    For me, I get a change in height I get 0.39739 m ~ 0.4 m heh. However, this is not the case since the pressure goes down as the spring is being pushed up =).
     
    Last edited: Jan 28, 2007
  20. Jan 29, 2007 #19
    heres what I get p2=1.64atm--its tricky because of the units always changing, but the A^2/k term needs to be multiplied by a million to convert from m^3 to liters. since 1.64 atm =98,690 N/M^2 *1.64,
    x=.81m.

    This result I like cuz it shows that the expansion is clearly constrained by the spring, and that the pressure is higher than if unconstrained.
     
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