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Homework Help: Rate of change of pressure with volume, adiabatic vs isothermal conditions

  1. Aug 13, 2010 #1
    1. The problem statement, all variables and given/known data
    For a particular sample of monatomic gas, the rate of change of pressure with volume under isothermal conditions is -2 x 10^7 Pa m^-3. Calculate the rate of change under adiabatic conditions.

    2. Relevant equations
    isothermal:
    dP/dV = -2 x 10^7
    Q = nRTln(Vf/Vi)

    monatomic:
    PV^(5/3) = constant

    adiabatic:
    dU = dW

    3. The attempt at a solution
    I feel like this question should be easy but I just don't even know how to approach it. I'd be extreme grateful if anybody could point me in the right direction. Thanks.
     
  2. jcsd
  3. Aug 13, 2010 #2

    Andrew Mason

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    Homework Helper

    1. What is dP/dV in terms of nRT and V for the isothermal process?

    2. What is the relationship between P and V in an adiabatic (reversible) process? (hints: it does not involve T and you need to know that it is a monatomic gas).

    3. Find dP/dV for the adiabatic process by differentiating the equation in 2.

    4. Compare the answers in 1. and 3. using the value for dP/dV that is given for the isothermal process.

    AM
     
  4. Aug 13, 2010 #3
    Thanks for your help. This is what I have so far:

    isothermal:
    PV = nRT
    => P = (nRT)/V
    => dP/dV = -(nRT)/(V^2)

    dP/dV = -2 x 10^7

    => -(nRT)/(V^2) = -2 x 10^7

    adiabatic:
    PV^(5/3) = k
    where k is some constant

    P = k/(V^(5/3))

    => dP/dV = -(5/3)k/(V^(8/3))

    I see that if I had values for n, T, and k I could work out the volume and use that to calculate a value for dP/dV in the adiabatic case -- but since I don't have those values and I'm not sure how to find them I'm afraid I'm still a little stuck.
     
  5. Aug 14, 2010 #4

    Andrew Mason

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    Science Advisor
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    Well, you know that:

    [tex]k = PV^\gamma[/tex]

    and PV = nRT

    and

    [tex]nRT/V^2 = 2 \times 10^7[/tex]

    AM
     
  6. Aug 17, 2010 #5
    Thanks for all your help. I've got it now.
     
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