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Homework Help: Gas inside an expandable cylinder raises lid on spring

  1. Sep 3, 2016 #1
    1. The problem statement, all variables and given/known data
    An expandable cylinder has its top connected to a spring of constant 2000 N/m. The cylinder is filled with 5L of gas with the spring relaxed at a pressure of 1 atmosphere and a temperature of 20C. If the lid has a cross-sectional area of 0.01m2 and negligible mass, how high will the lid rise when the temperature is raised to to 250C?

    2. Relevant equations
    PV = nRT
    PV/T initial = PV/T final

    3. The attempt at a solution
    P = Force/Area. When the gas expands it pushes on the spring which pushes back with F = k * displacement. Displacement here is h so P=kh/A. We already know the initial pressure is 1 atmosphere, so the final P is kh/A + 1.

    The initial volume is 5L, so the final volume must be 5 + Ah.

    The initial temperature and final temperature are given as 20 and 250C.

    Plugging that in gives:

    (1*5)/20 = (1 + kh/A)(5 + Ah)/250

    62.5 = (1 + kh/A)(5 + Ah)

    which yields a quadratic:

    62.5 = 5 + Ah + 5kh/A + kh2

    Ah = .01h and 5kh/A = 106h so Ah can be neglected when adding these terms ≈ 106h.

    0 = -57.5 + 106h + 2000h2

    When I solve this quadratic equation I get h = 5.75E-5. But this isn't the right answer. I'm stuck on figuring out where I went wrong so help is very appreciated. Thank you!
  2. jcsd
  3. Sep 3, 2016 #2


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    I think you are setting it up correctly. But you need to watch your units. Plug in units along with the numbers to make sure the terms in your equations are consistent in units.
  4. Sep 3, 2016 #3
    If I write out units I get

    (1 atm)(5 L)/(20 C) = (1 atm + kh/A m-1)(5 L + Ah m2)/(250 C)

    .25 atm*L/C = (5 atm*L + 5kh/A L*m-1 + kh2m2)/(250 C)

    (.25 atm*L/C)*(250 C) = 62.5 atm*L = 5 atm*L + 106h/A L*m-1 + 2000h2m2)

    57.5 atm*L = 106h/A L*m-1 + 2000h2m2)

    L can be converted into m3:

    (57.5 atm*L)(.001m3/L) = 106h/A L*m-1*(.001m3/L) + 2000h2m2)

    .0575 atm*m3= 4hm2 + 8h2m2

    But I am not sure what to do now that I see the units on either side are different.
  5. Sep 3, 2016 #4


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    (PV/T)initial = (PV/T)final is a result of the ideal gas law PV = nRT. What type of temperature scale is required for T here?

    The spring constant is given in N/m which are units belonging to the SI system of units. Are the liter and the atmosphere SI units for volume and pressure?
  6. Sep 3, 2016 #5
    Ok, so now I have:

    (1.01E5 N*m-2)(.005 m3)/(293 K) = (1.01E5 N*m-2 + kh/A N*m-2)(.005 m3 + Ah m3)/(523 K)

    (523/293)(.005)(1.01E5 N*m) = (1.01E5 N*m-2 + kh/A N*m-2)(.005 m3 + Ah m3)

    But that simplifies to:

    (523/293)(.005)(1.01E5 N*m) = (.005)(1.01E5) N*m + (.01)(1.01E5)h N*m + (.005)(2000)/(.01)h N*m + (.01)(2000)/(.01)h2 N*m

    which is in N*m and it should be in meters.
  7. Sep 3, 2016 #6


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    That looks good. Note that you have shown that each term has total units of N*m. So, your equation is consistent in terms of units. That doesn't mean that h has units of N*m. You have already taken account of the fact that h has units of meters in order to obtain units of N*m for the terms that contain h. So, in your way of writing the equation, h is now just a number without units since you have already included the unit for h when writing N*m for the entire term.

    This is not the usual way to write it. The usual way to write, say, the term (.01)(1.01E5)h with units would be

    (.01 m2)(1.01 x 105 N/m2)h = [(.01)(1.01 x 105) N]h = (1.01 x 103 N)h.

    The units for h are not written explicitly here, but it is assumed that h carries the units of meters.
    Overall, the expression (1.01 x 103 N)h still has units of N*m, where the m comes from the units of h.
  8. Sep 4, 2016 #7
    When I plugged in everything and solved the quadratic I got .169m, which is the correct answer. I have always had trouble with units so this was very helpful, thank you!
  9. Sep 4, 2016 #8


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    OK, good work!
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