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The first Law of Thermodynamics seems confusing to me.

  1. Sep 11, 2011 #1
    Hi,

    The first law of thermodynamic is sometimes written as
    [itex]\Delta[/itex]U = [itex]\Delta[/itex]Q-P [itex]\Delta[/itex]V and sometimes as [itex]\Delta[/itex]U = Q+W, and sometimes [itex]\Delta[/itex]U = Q - W. I am confused about all these.
    I also know that W is positive when the system in question does work to its surrounding and negative otherwise. Looking at [itex]\Delta[/itex]U = [itex]\Delta[/itex]Q-P [itex]\Delta[/itex]V, i want to conclude that every expansion entails positive work and every compression negative work. I'm I correct to do so?
     
  2. jcsd
  3. Sep 11, 2011 #2
    Yep, that's right.
     
  4. Sep 11, 2011 #3
  5. Sep 12, 2011 #4
    Thanks to all for your help. I am actually grabbing some sense out of your comments.
    My problem still remains.
    Let's consider for instance an Adiabatic expansion of an ideal gas. We have the first law:
    [itex]\Delta[/itex]U = Q - W.
    Now Adiabatic process [itex]\Rightarrow[/itex] Q = 0, hence
    [itex]\Delta[/itex]U = - W.
    But W = P [itex]\Delta[/itex]V [itex]\succ[/itex] 0 for expansion; assuming constant P.
    This Would mean [itex]\Delta[/itex]U [itex]\prec[/itex] 0 i.e Internal energy decreases, when the gas expands at constant pressure.
    Now considering the equation of state of an ideal gas
    U = [itex]\frac{3}{2}[/itex] PV [itex]\Rightarrow[/itex] [itex]\Delta[/itex]U =[itex]\frac{3}{2}[/itex] P[itex]\Delta[/itex]V.
    Expansion would mean [itex]\Delta[/itex]V [itex]\succ[/itex] 0, hence
    [itex]\Delta[/itex]U [itex]\succ[/itex] 0 , so that the internal energy increases in this case.

    Is there something i'm getting wrong in the above reasoning?
    Thanks for helping.
     
  6. Sep 12, 2011 #5
    You cannot use PV=nRT directly for an adiabatic expansion since temperature is not constant so you cannot integrate the first law.

    For adiabatic first law

    dq=0; dU=dw=-PdV

    dw = CvdT

    -PdV = CvdT

    [tex]{C_v}dT + PdV = 0[/tex]

    [tex]{C_v}\frac{{dT}}{T} + nR\frac{{dV}}{V} = 0[/tex]

    integrating between initial and final temps and initial and final volumes

    [tex]{C_v}\ln \frac{{{T_2}}}{{{T_1}}} + nR\ln \frac{{{V_2}}}{{{V_1}}} = 0[/tex]

    if [tex]\gamma = \frac{{{C_p}}}{{{C_v}}}[/tex]

    and nR = Cp - Cv

    then we can arrive at

    [tex]\frac{{{T_1}}}{{{T_2}}} = {\left( {\frac{{{V_2}}}{{{V_1}}}} \right)^{\gamma - 1}}[/tex]

    and

    PV[itex]\gamma[/itex] = constant
     
    Last edited: Sep 12, 2011
  7. Sep 12, 2011 #6
    It seems that you are mixing the adiabatic and constant pressure processes.
    In the adiabatic expansion the internal energy decreases. The pressure is not constant so you cannot write the change in U as 3/2 P*Delta V.

    For constant pressure expansion heat must be transferred to the gas. The internal energy of the ideal gas will increase in this case.
     
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