# Deriving the formula for the change in entropy

1. May 26, 2014

### 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 = 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. May 26, 2014

### 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.

3. May 26, 2014

### 2sin54

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.

4. May 26, 2014

### 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.

5. May 26, 2014

### WannabeNewton

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.