Show that the entropy is non-negative

[patrickmoloney]

Homework Statement

Two vessels A and B each contain N 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 P and at temperatures T_A and T_B. The two vessels are now brought into thermal contact, with the pressure of the gases kept constant at the value P

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

Homework Equations

Q_{in} = -Q_{out}

\Delta S = \int \dfrac{dQ}{T}

The Attempt at a Solution

\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{)}
\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{)}

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

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

 

[Chestermiller]

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

 

[patrickmoloney]

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

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

Edit: However

T_f = \Big{(}\dfrac{T_A +T_B}{2}\Big{)} 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.