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.