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Adding heat to a Non Ideal Gas problem

  1. Sep 14, 2014 #1
    1. The problem statement, all variables and given/known data

    When 120J of heat are added to this gas, the temperature of the gas changes. Find the change in temperature if the heat is added at (a) constant volume. (b) constant pressure.



    2. Relevant equations
    The equation of state of at the gas is (p + 50)V = 10T

    The internal energy of this gas is given by U = 20T + 50V + 40


    3. The attempt at a solution

    For part (a) I used the fact that U = W + Q, but since the volume is constant I know that U=Q=120J. From this I substitute V for the equations to obtain an equation for T with pressure as the independent variable. T = (4p + 200)/(75 + p) then dT/dp = 100/(75+p)^2. Is this the right way to go about this problem? or should I use something like since U = Q and the differential of U(T,V) is
    dU = (partial dU/dT)DT + (partial dU/dV)DV and then take the partial of U with respect to T, which would be dU = 20 dT. THEN 20 dT = 120J since U = Q. Thus, dT = 6. I am not sure which route to go. Similar question for part (b). Thanks for any advice!
     
  2. jcsd
  3. Sep 14, 2014 #2
    For part a, your second method is correct. You are looking for the changes in U and T (not their absolute values), which this method captures.

    Let's see what you are able to come up with for part b.

    Chet
     
  4. Sep 15, 2014 #3
    Thanks Chet! for part (b) I know that the pressure is constant so U = 120J + W = 120J + PV. Then we can get 120J + PV = 20T + 50V + 40J then this simplifies to 80J + PV = 20T +50V. Since from the equation of state PV = 10T - 50V we can substitute and get 80J + 10T - 50V = 20T + 50V. This can be rearranged to
    10T = 80J - 100V then taking the differential would give 10dT= -100 which means dT = -10 K. What do you think?
     
  5. Sep 15, 2014 #4
    I think it's not right. First, solve the equation of state for V as a function of p and T. Then determine ΔV as a function of ΔT, holding p constant. The amount of heat is:

    Q = ΔU+pΔV

    Express both terms on the right by eliminating ΔV. The pressure p should cancel, and you should end up with just a function of ΔT.

    Chet
     
  6. Sep 15, 2014 #5
    Okay, so solving for my equation of state for V, I get V = 10T/(p+50), then this means dV = 10dT/(p+50) since p is constant. Also, I know that Q = du + dw and du = 20dT +50dV = 120J - pdV. This can be rewritten as 20dT +dV(p + 50) = 120J. Then substituting the above dV I get 30dT = 120. Thus, dT = 4K.
     
  7. Sep 15, 2014 #6
    Good job.

    Chet
     
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