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Some Thermo questions

  1. Sep 15, 2005 #1
    There are a couple of questions on my practice exam I am stuck on. If you could point me in the right direction I'd appreciate it.

    1. A gas is described by the equation PV= RT+aT^2 where a is some gas specific constant. Find (dU/dV)@constant T

    And provide an expression for the isothermal reversible work

    I think I was able to solve for the work as it is just the integral of pdV which is easy enough to compute. If this is incorrect could someone please tell me.


    2. The minimum amount of work to cool a refrigerator from 300K to 200K if the heat capacity of the refrigerator is 1000J/K. The exterior is assumed to be at a constant 300K despite being used as a heat sink.

    The hint for this one says to think of it as a carnot engine in reverse but I don't know what this means exactly. I found the efficiency of this if it were a Carnot engine and it is 1/3 then I just assumed minimum work was the product of the efficiency heat capactiy and temperature difference (300-200) . I suspect however that this is wrong.

    3. Lastly, the heat capacity of some monatomic ideal gas is Cv=2.5R and Cp=3.5R, what is the heat capacity of a process where P/V is a constant i.e. the ratio of P:V is constant. I gathered that if P/V is constant, as PV=nRT then the process is isothermal and that perhaps the Cp/v is infinite, as dU for an isothermal process is 0.
     
    Last edited: Sep 15, 2005
  2. jcsd
  3. Sep 15, 2005 #2
    self-bump

    also in retrospect the last problem does not actually constitute an isothermic process
     
  4. Sep 16, 2005 #3

    LeonhardEuler

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    I think you did the first problem correctly. For the second:
    [tex]\eta_{th}=\frac{w}{q_h}[/tex]
    [tex]\eta_{th}=1-\frac{T_c}{T_h}[/tex]
    Therefore:
    [tex]w=(1-\frac{T_c}{T_h})q_h[/tex]
    I hope that is enough of a hint.

    For the last one, I guess that by "heat capacity for the process" they mean the heat supplied divided by the change in T. Consider a process hapening in two steps: one at constant volume and the next at constant P. Add the dq's and divide by the dT's.
     
  5. Sep 16, 2005 #4

    SpaceTiger

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    I may be missing something, but this problem appears to be non-trivial. If you're working out of Reif, then I suggest looking at page 154. If not, consider the following relations:

    [tex]dU=(\frac{\partial U}{\partial T})_VdT+(\frac{\partial U}{\partial V})_TdV[/tex]

    [tex]TdS=dU+pdV[/tex]

    Substituting the expression for p and dU, you then need to solve for

    [tex]\frac{\partial^2S}{\partial V \partial T}=\frac{\partial^2S}{\partial T \partial V}[/tex]

    Remembering that the second derivatives will be the same regardless of order of differentiation. The expression you're looking for will eventually drop out.


    I suspect your answer is incorrect. If you did the following:

    [tex]W=\int pdV=\int \frac{RT+\frac{1}{2}aT^2}{V}dV[/tex]

    This will be a trivial integral only if T is constant, but reversible work implies constant entropy, not temperature. I suspect you'll have to rederive the expressions for adiabatic expansion/compression.


    I have to go to class soon and I won't be back online till tomorrow, but one more quick thing. If P/V is constant, then

    [tex]\frac{P}{V}=\frac{nRT}{V^2}=constant[/tex]

    which does not imply isothermality.
     
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