Deriving the formula for the change in entropy

[2sin54]

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 = nC_vdT which in itself is the equation for unit of heat transferred (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

 

[WannabeNewton]

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.

 

[2sin54]

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.

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.

 

[CAF123]

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.

 

[WannabeNewton]

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.

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.