# Entropy of a mole of an ideal gas

1. Jun 7, 2009

### fluidistic

1. The problem statement, all variables and given/known data

Show that the entropy of one mole of an ideal gas is given by $$\Delta S = \int \frac{C_v dT+PdV}{T}$$.

2. The attempt at a solution
$$\Delta S= \frac{dQ}{T}$$. From the 1st of Thermodynamics, $$\Delta U= \Delta Q - \Delta W$$, where $$W$$ is the work done by the gas.

Hence $$\Delta Q = \Delta U + \Delta W \Rightarrow dQ=dU+dW=C_vdT+PdV$$. So $$\Delta S= \int \frac{C_vdT+PdV}{T}$$.$$\square$$.
But this is cheat. By this I mean that I didn't know that $$dU=C_vdT$$. I deduced it because I had to fall over the result. How can I deduce $$dU=C_vdT$$, for an ideal gas?
Where did I supposed the 1 mol of the ideal gas?
Last question, what are the bounds of the integral of the change of entropy?
Because $$\Delta S = \int \frac{C_v dT+PdV}{T}=\int \frac{C_vdT}{T}+ \int \frac{PdV}{T}$$ and I'm sure the bounds of the 2 integrals are different. For instance I think that the bounds of $$\int \frac{PdV}{T}$$ are $$\int_{V_1}^{V_2} \frac{PdV}{T}$$. And I guess that the bounds of $$\int \frac{C_vdT}{T}$$ are $$\int_{T_1}^{T_2} \frac{C_vdT}{T}$$ but I'm not sure.

2. Jun 7, 2009

### Mapes

By definition,

$$C_X=T\left(\frac{\partial S}{\partial T}\right)_X$$

where $X$ is some constraint. This is easily combined with $dU=T\,dS-P\,dV$.

If you define $S$ and $V$ as the molar entropy and molar volume, respectively, then everything works out. You'll just multiply the answer by 1 mole.

I don't believe it's possible to write a definite integral that contains both $dT$ and $dV$. Combining them is a useful shorthand that works for the indefinite integral only.

3. Jun 7, 2009

### fluidistic

Ok, thank you very much!

4. Jun 7, 2009

### Andrew Mason

For a constant volume process, W = 0. From the first law, if W = 0, dQ = dU.

By definition dQ = nCvdT for a constant volume process. So dU = nCvdT.

AM

5. Jun 7, 2009

### fluidistic

Thanks. I think I understand well now.