Show that the entropy is non-negative

[patrickmoloney]

Homework Statement

Two vessels [itex]A[/itex] and [itex]B[/itex] each contain [itex]N[/itex] molecules of the same perfect monatomic gas. Initially, the two vessels are thermally isolated from each other, with the two gases at the same pressure [itex]P[/itex] and at temperatures [itex]T_A[/itex] and [itex]T_B[/itex]. The two vessels are now brought into thermal contact, with the pressure of the gases kept constant at the value [itex]P[/itex]

Find the change in entropy after equilibrium and show that this change is non-negative.

Homework Equations

[tex]Q_{in} = -Q_{out}[/tex]

[tex]\Delta S = \int \dfrac{dQ}{T}[/tex]

The Attempt at a Solution

[tex]\Delta S_A = \int_A^f \dfrac{dQ_A}{T}= \int_A^f \dfrac{mcdT}{T}= mc \ln \Big{(}\dfrac{T_f}{T_A}\Big{)}[/tex]
[tex]\Delta S_B = \int_B^f \dfrac{dQ_B}{T}= \int_B^f \dfrac{mcdT}{T}= mc \ln \Big{(}\dfrac{T_f}{T_B}\Big{)}[/tex]

[tex]\Delta S_{tot} = \Delta S_B - \Delta S_A = mc \ln \Big{(}\dfrac{T_A}{T_B} \Big{)}[/tex]

how can a prove that [itex]\ln \Big{(}\dfrac{T_A}{T_B}\Big{)} > 0[/itex]

 

[Chestermiller]

It should be the summation of the entropy changes, not the difference.

 

[patrickmoloney]

Oh yeah! Thanks [tex]\Delta S_{tot}= mc \ln \Bigg{(}\dfrac{T_{f}^2}{T_A T_B}\Bigg{)}[/tex]

I know that [itex]mc > 0[/itex]. How do I prove that [tex]\dfrac{T_{f}^2}{T_A T_B} >1[/tex] I suppose [itex]T_{f}^2 \gg T_A T_B[/itex] but that's not a proof.

Edit: However

[tex]T_f = \Big{(}\dfrac{T_A +T_B}{2}\Big{)}[/tex] so I guess that's it.

 

[Chestermiller]

You need to determine Tf using the 1st law.

 

[patrickmoloney]

You need to determine Tf using the 1st law.

Yeah I thought so! Thanks a lot.