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Two questions on ideal gases and heat

  1. Nov 10, 2012 #1
    1. The problem statement, all variables and given/known data

    A horizontal cilindre with a piston of mass M = 0.5 kg is filled with air (the specific heat of air is Cp = 1000 Joule/Grad*kmole). The heating of the gas results in the piston's accelerated displacement (with constant acceleration) until the velocity v = 1 m/s. Determine the amount of heat acumulated by the gas during the heating. The energy of one gas mol is U = Cp*T. Ignore the external pressure and the thermical capacities of both the cilindre and piston. The universal gas constant is R = 8.31*10³ Joule/Grad*kmole.




    The volume of a room is 100m³ with ambient pressure of 1.02*10^5 Pa. Determine the air mass which left the room after the raise in temperature from 10°C to 25°C. The air density in normal conditions (To = 273K, p0 = 1.01*10^5 Pa) is 1.29kg/m³.



    2. Relevant equations

    General gas equations, minding the circumstances:
    P1V1 = P2V2
    P1/T1 = P2/T2
    V1/T1 = V2/T2
    PV = nRT

    Energy conservation
    dU = dQ - dW

    3. The attempt at a solution

    I guess the solutions comes out rather simply once you know the conditions. The problem is I don't exactly know how the various factors (pressure, volume, density) are linked, how they affect each other. And thus I have no ground for assumptions. Could someone explain these for me? Thanks!
     
    Last edited: Nov 10, 2012
  2. jcsd
  3. Nov 10, 2012 #2

    Simon Bridge

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    Consider the different parameters: P,V,T : are any of them constant?
     
  4. Nov 10, 2012 #3
    The work done is W = ∫F dx = ∫m a dx = m∫a dx (from 0 to distance x)

    a = dv/dt = (dv/dx)(dx/dt) = (dv/dx)v → a dx = v dv
    → ∫a dx (from 0 to distance x) = ∫v dv (from 0 to 1 m/s).

    Thus, W = m∫v dv (from 0 to 1 m/s) = mv^2 /2

    Assuming constant pressure (since you provided Cp for air), we have that the work done is:

    W = ∫P dV = P∫dV = PV2 - PV1 = nRT2 - nRT1 = nRΔT. Hence, ΔT = W/nR.

    Thus, the amount of heat absorbed by the gas = U + W = CpΔT + W = W(1 + Cp/(nR)) =

    = mv^2(1+Cp/(nR))/2

    Assuming only one mol (since you provided U for 1 mol):

    Q = mv^2(1+Cp/(1 mol * R))/2
     
  5. Nov 10, 2012 #4

    Simon Bridge

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    @LoadedAnvils:... usually we like to encourage people to do their own homework: the idea is that people learn best by doing the work themselves. What we normally do is try to help them get unstuck.
     
  6. Nov 11, 2012 #5
    Oh, thanks LoadedAnvils! Our professor tries to avoid the use of calculus in questions he himself makes, so I didn't even thought of that. Well, guess it was needed this time! Thanks!

    Simon Bridge, I think you can consider the pressure being constant, right? I was a little lost because I thought the room was closed, and then someone goes and, I don't know, drill a hole on the wall to make the air come out.
    So, since pressure is constant, we have

    V1/T1 = V2/T2
    V2 = (100*298)/283 = 105,3 m³

    or, 5,3 m³ of air left the room.
    Now we have to find the air density, say d2, under the new conditions. In the normal conditions, we have

    d = m/V0 = 1,29 kg/m³

    and so

    V0 = m/1,29

    Using Clayperon's equation for the final condition and the normal condition,

    P0V0 = nRT0
    P2V2 = nRT2

    or

    V2 = (P0*V0*T2)/(P2*T0)

    And thus

    d2 = m/V2 = (m*P2*T0)/(P0*V0*T2) = (d*P2*T0)/(P0*T2) = 1,193 < 1,29

    So the mass which left the room is 5,3 * 1,193 = 6,32kg.
    Could someone confirm this result for me? Thanks!
     
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