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Homework Help: Ideal gas in a cylinder with a piston

  1. Apr 21, 2017 #1
    1. The problem statement, all variables and given/known data
    A vertical cylinder of radius r contains an ideal gas and is fitted with a piston of mass m that is free to move. The piston and the walls of the cylinder are frictionless, and the entire cylinder is placed in a constant-temperature bath. The outside air pressure is p0. In equilibrium, the piston sits at a height h above the bottom of the cylinder. (a) Find the absolute pressure of the gas trapped below the piston when in equilibrium. (b) The piston is pulled up by a small distance and released. Find the net force acting on the piston when its base is a distance h + y above the bottom of the cylinder, where y << h.

    2. Relevant equations
    F = pA = ma
    pV = nRT (pressure p, volume V, moles in gas n, gas constant R, temperature T)
    V = πr2h

    3. The attempt at a solution
    I managed to solve a) by myself and get p1 = p0 + mg/πr2

    Trying to solve b)
    Positive force is upwards so ∑F = F - F0 - Fp where F is the force upwards on the piston by the pressure of the gas underneath, F0 is the downward force on the piston by the air pressure p0 above and Fp is the downward force by the weight of the piston.

    F0 = p0A = p0πr2
    Fp = mg

    moles in ideal gas n = p1V1/RT1 (1)
    pressure p2 = nRT2/V2 (2)
    Because of heat conduction and the constant temperature outside the cylinder, T2 = T1. Inserting (1) in (2) yields:
    p2 = p1V1/V2
    V1 = πr2h
    V2 = πr2(h + y)
    p2 = p1h/(h + y)

    ∑F = p2A - p0A - mg. A = πr2.
    Simplifying yields my answer: ∑F = (h/(h + y) - 1)(mg + p0πr2)
    Textbook answer: ∑F = -(y/h)(mg + p0πr2)

    I tried replacing the variables with values to see the result of both answers. My answer is not correct apparently, what did I do wrong?
  2. jcsd
  3. Apr 21, 2017 #2


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    Homework Helper

    Your answer is correct, but simplify the factor h/(h+y)-1, and use that y<<h.
  4. Apr 21, 2017 #3
    I don't even know what y<<h is. And simplifying h/(h+y)-1 gives me (h-y)/(h+y)
  5. Apr 21, 2017 #4
    Check your math. This is wrong.
  6. Apr 21, 2017 #5
    h/(h+y) - 1 = h/(h+y) - (h+y)/(h+y) = (h - (h+y))/(h+y) = (h - h - y)/(h+y) = -y/(h+y)
    oops, you're right. But what am I supposed to do now?
  7. Apr 21, 2017 #6
    Dividing numerator and denominator by h yields:
    $$\frac{y}{h+y}=\frac{y}{h}\frac{1}{(1+\frac{y}{h})}$$For y much smaller than h, y/h is much smaller than 1. Therefore, $$\frac{1}{(1+\frac{y}{h})}\rightarrow 1$$and $$\frac{y}{h+y}\rightarrow \frac{y}{h}$$
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