1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Isothermal Compression of a Ideal Gas and Distance

  1. Jun 13, 2016 #1
    1. The problem statement, all variables and given/known data
    The figure at the bottom shows a cylindrical tank of diameter D with a moveable 3.00-kg circular disk sitting on top. The disk seals in the gas inside but is able to move without friction. The gas inside is at temperature T. The height of the disk is initially at h = 4.00 m above the bottom of the tank. When a lead brick of mass 9.00 kg is gently placed on the disk, it moves downward, compressing the gas and decreasing the volume. Assuming the temperature of the gas is held constant, at what distance from the bottom of the will the disk come to rest?

    media%2F38a%2F38ae8a2b-f40e-4de5-a308-1866746910bb%2Fphp3V4qiK.png

    mass_{disk}=3.00 kg
    mass_{Pb}=9.00 kg
    m_{T}=12.00kg
    h_i=4.00m
    Isothermal \therefore T_i=T_f
    V_i > V_f ; P_i<P_f

    2. Relevant equations
    \P = \rho*\g*\h
    P_1*V_1*T_2=P_2*V_2*T_1
    W=nRT*ln(\frac{V_f}{V_i})=m*g*d

    3. The attempt at a solution

    \
    delta P = -\rho*g*\int_{y}^{4} = -0.5*\rho*g*(y^2-16)
    F_{disk+Pb}=(m_{d}+m{Pb})/g=127.53N
    P_1*V_1=P_2*V_2
    P_1*A_1*H_1=P_2*A_2*H_2 (A_1=A_2)
    H_2=4*\frac{P_1}{P_2}

    I'm not really sure where to proceed here since the initial pressure nor the final pressure nor any of the volumes are given so i'm generally extremely confused on how to proceed to find the final distance the piston would rest at. I know since that by Boyles law as Volume decreases, Pressure increases at constant Temperature. I'm just not sure how we factor any of that in without being given volume or area.
     

    Attached Files:

  2. jcsd
  3. Jun 14, 2016 #2

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    I haven't spent the time to work out the solution but I note that Pressure = Force/Area so my guess would be that the area cancels at some point.
     
  4. Jun 14, 2016 #3
    Algebraically, in terms of mdisk and g, what is the force of the gas on the disk initially? Algebraically, in terms of mdisk, g, and A (where A is the area of the disk), what is the pressure of the gas P1 on the disk initially?

    Algebraically, in terms of mdisk, mPb, and g, what is the force of the gas on the disk finally? Algebraically, in terms of mdisk, mPb, g, and A (where A is the area of the disk), what is the pressure of the gas P2 on the disk finally?

    Algebraically, what is the ratio of P1 to P2?
     
  5. Jun 14, 2016 #4
    I've got this but I'm not sure how to proceed from here without being given any of the pressures.

    $$\delta P = -\rho*g*\int_{y}^{4} = -0.5*\rho*g*(y^2-16)$$
    $$ F_{disk+Pb}=(m_{d}+m{Pb})/g=127.53N $$
    $$ P_1*V_1=P_2*V_2 $$
    $$ P_1*A_1*H_1=P_2*A_2*H_2 (A_1=A_2) $$
    $$H_2=4*\frac{P_1}{P_2}$$
     
  6. Jun 14, 2016 #5
    You haven't answered my questions.
     
  7. Jun 14, 2016 #6
    So the initial Force is described algebraically as
    Fg=mdisk*g
    Initial Pressure
    P1=F/A=mdisk*g/A
    P2=mdisk+Pb*g/A
    P1/P2=V2/V1=A*H2/A*H1

    mdisk*g*A-1/mdisk+Pb*g*A-1= 4(mDisk/mPb+Disk)=H2

    Is that how height gets related to pressure? What about atmospheric pressure or does that not factor in?
     
  8. Jun 14, 2016 #7
    VERY NICELY DONE!!!

    The analysis you have done so far assumes that the outside pressure is zero (i.e., vacuum). Let's go back to the equilibrium force balances on the piston, and repeat them assuming that the outside pressure is atmospheric (patm). Again, please do this algebraically. What do you get for the inside gas pressures then?
     
  9. Jun 14, 2016 #8
    P1=F/A=mdisk*g/A + Patm
    P2=mdisk+Pb*g/A + Patm
    P1/P2=mdisk*g/A + Patmmdisk+Pb*g/A + Patm


    The areas no longer cancel out here so im guessing the atmospheric pressures cancel somehow?
     
  10. Jun 14, 2016 #9
    No. I guess they expected you to assume vacuum outside the cylinder. But you did great on the case with atmospheric pressure, especially realizing that you need to know A to get a solution.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Isothermal Compression of a Ideal Gas and Distance
Loading...