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Thermodynamics process

  1. Jun 12, 2007 #1
    1. The problem statement, all variables and given/known data

    i have a thermdyanmics process. with a constant pressure, constant volume and adiabatic. please see the attachment.

    2. Relevant equations

    from C->A (constant pressure)
    so change in pressure = 0
    equations needed to do this problem:


    Pressure* V = n *R* T -----------> main equation
    heat = Q = Cp * n * change in temperature ----------------> main
    (Cp is called the specific heat at constant pressure)

    3. The attempt at a solution

    our goal is use Q = Cp n * change in temperature

    BUT we don't know change in temperature
    we know Cp = 20.8 for an ideal monatomic gas (just a constant you look up in a physics textbook)
    n = 1 mole

    so how do we find change in temperature?

    we need to use P Vinitial = n R Tinitial
    400 * (1) = 1 * 8.315 * Tinitial

    so Tinitial = 48.1058 Kelvin

    P Vfinal = n R Tfinal

    400 * 2 = 1 * 8.315 * Tfinal
    so Tfinal = 96.2117 Kelvin

    now we are done!

    Q = n * Cp * change in temperature

    Q = (1) *( 20.8 J/mol*K ) (96.2117 - 48.1058 Kelvin)
    Q = 1000.6 Joules


    is this right?

    thank you
     

    Attached Files:

  2. jcsd
  3. Jun 12, 2007 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    This is just a first law problem: dQ = dU + dW where dW is the work done by the gas. For C to A:

    (1) [tex]\Delta Q_{CA} = \Delta U_{CA} + W_{CA} = \Delta U_{CA} + P\Delta V_{CA}[/tex]

    You know that the change internal energy from C to A is +100 J (to balance the 200 J loss from A to B and 100 J gain from B to C).

    You can also determine the work done from C to A because you are given the pressure and change in volume.

    So plug those values in (1) to give you [itex]\Delta Q_{CA}[/itex]

    AM
     
  4. Jun 12, 2007 #3
    thank you!
    why was i off? i thought i had this!


    so Q = U + W
    Q = 100 + (400pascals * (2-1)m^3)
    Q= 500 joules?

    or is it

    Q = 100 - (400pascals * (2-1)m^3)
    Q= -300 joules.

    which one is right?

    thank you!
     
  5. Jun 12, 2007 #4

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    Heat flow out of the gas is negative. Heat flow into the gas is positive. Work done by the gas is positive. Work done on the gas is negative.

    So:

    [tex]\Delta Q = \Delta U + P\Delta V = +100 + 400 * (-1) = -300 J[/tex]

    This means that 300 J of heat flows out of the gas in going from C to A.

    AM
     
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