- #1

fluidistic

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## Homework Statement

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

**2. The attempt at a solution**

[tex]\Delta S= \frac{dQ}{T}[/tex]. From the 1st of Thermodynamics, [tex]\Delta U= \Delta Q - \Delta W[/tex], where [tex]W[/tex] is the work done by the gas.

Hence [tex]\Delta Q = \Delta U + \Delta W \Rightarrow dQ=dU+dW=C_vdT+PdV[/tex]. So [tex]\Delta S= \int \frac{C_vdT+PdV}{T}[/tex].[tex]\square[/tex].

But this is cheat. By this I mean that I didn't know that [tex]dU=C_vdT[/tex]. I deduced it because I had to fall over the result. How can I deduce [tex]dU=C_vdT[/tex], 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 [tex]\Delta S = \int \frac{C_v dT+PdV}{T}=\int \frac{C_vdT}{T}+ \int \frac{PdV}{T}[/tex] and I'm sure the bounds of the 2 integrals are different. For instance I think that the bounds of [tex]\int \frac{PdV}{T}[/tex] are [tex]\int_{V_1}^{V_2} \frac{PdV}{T}[/tex]. And I guess that the bounds of [tex]\int \frac{C_vdT}{T}[/tex] are [tex]\int_{T_1}^{T_2} \frac{C_vdT}{T}[/tex] but I'm not sure.