Entropy of a mole of an ideal gas

[fluidistic]

Homework Statement

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.

 

[Mapes]

How can I deduce dU=C_vdT, for an ideal gas?

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.

Where did I supposed the 1 mol of the ideal gas?

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.

Last question, what are the bounds of the integral of the change of entropy?

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.

 

[fluidistic]

Ok, thank you very much!

 

[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

 

[fluidistic]

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

Thanks. I think I understand well now.