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Deriving the formula for the change in entropy

  1. May 26, 2014 #1
    Hello. I was reading Hyperphysics website and could not get one particular part. I am providing a picture of the equation I am having trouble with: http://i.snag.gy/W3CC3.jpg
    The particular part that puzzles me is the relation around the third equation sign. From the formula there one can think that dU = nCvdT which in itself is the equation for unit of heat transfered (dQ). Obviously it would be correct if there was no work done (dV = 0) but it is not the case here. What am I not seeing here? The full link is http://hyperphysics.phy-astr.gsu.edu/hbase/therm/entropgas.html
     
  2. jcsd
  3. May 26, 2014 #2

    WannabeNewton

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    Actually ##dV = 0## is the case as far as the equality ##dU = nC_V dT## is concerned because ##nC_V = (\frac{\partial U}{\partial T})_V##. Remember ##C_V## is the specific heat at constant volume.
     
  4. May 26, 2014 #3
    Hmm. That makes sense, yes. But what still is strange to me is that they use the first law of thermodynamics ( dQ = dU + dA ) to express dQ, then they leave dA without touching it and express dU using the same law again, just in different conditions (dV = 0). So in the end there are *two works* one which is left as pdV while the other one is 0.
     
  5. May 26, 2014 #4

    CAF123

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    The first law of thermodynamics can be written, in differential form, as ##dU = dQ - pdV##, where -pdV is the work done by the surroundings as the system moves through a series of quasi-static states. Now take the partial derivative of both sides wrt T at constant V. The result is that $$\left(\frac{\partial U}{\partial T}\right)_V = \left(\frac{\partial Q}{\partial T}\right)_V - p \left(\frac{\partial V}{\partial T}\right)_V $$ The second term on the RHS is zero and the first term is ##C_V##
    U is a state variable, so you can always consider a constant volume process between the initial and final states. For an ideal gas U=U(T) so dU = C_V dT.
     
  6. May 26, 2014 #5

    WannabeNewton

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    There is only one process being done and it's the general one between two thermal equilibrium states of the system wherein ##dV \neq 0## in general and ##\delta W## (##= pdV## for quasi-static process) ##\neq 0## in general, as a result. The specific heat ##c_V## is a property of the system being considered and is entirely independent of the process we are performing on the system in going from one equilibrium state to another. Sure we calculate ##c_V## by taking our system and considering a constant volume process, so as to use ##c_V = \frac{1}{n}(\frac{\partial U}{\partial T})_V##, but this is a one-time deal as it is an inherent property of the system itself and therefore once we calculate ##c_V## we can then use it in calculations for any other process whatsoever taking the system from one equilibrium state to another, such as the general one considered above with non-zero work done by the system.
     
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