How to calculate the transition temperature in this problem?

[Sayantan21]

Summary: The transition Sn(s, gray) ⇌ Sn(s, white) is in equilibrium at 18°C and 1 atm pressure. If ΔS = 8.811K mol for the transition at 18°C and if the densities are 5.75 g/cm3 for gray tin and 7.28 g/cm3 for white tin, calculate the transition temperature under 100 atm pressure

The transition Sn(s, gray) ⇌ Sn(s, white) is in equilibrium at 18°C and 1 atm pressure. If ΔS = 8.811K mol for the transition at 18°C and if the densities are 5.75 g/cm3 for gray tin and 7.28 g/cm3 for white tin, calculate the transition temperature under 100 atm pressure

 

[mjc123]

You need to show an attempt of your own before receiving help, this is a forum rule.

There's no point repeating the same paragraph 3 times. And if you're going to, copy it right. ΔS = 8.8 J/K mol, not 8.811K mol.

What quantities determine the transition temperature? How might they vary with pressure?

 

[Sayantan21]

dp/dT = ΔS /ΔV
dp/dT = ΔS /ΔV

(P2-P1)/(T2-T1) = ΔS/(M/d1-M/d2); where d1 and d2 are densities of gray tin and white tin, M is equal to molecular weight of tin.
P2 = 100 atm ; P1= 1atm T2=? T1= 273+ 18 =291K ;
(100-1)/(T2-291) = 8.811/ (118.71/5.75 - 118.71/ 7.28)

 

[mjc123]

That looks OK in principle; I did it a bit differently, but it should come to the same answer. Just a few points you should be careful about:
- ΔS = 8.8, not 8.811, as I've already said.
- ΔV is Vwhite - Vgrey, make sure you get the sign right.
- Careful with units; what are the units of P? ΔV? ΔS? Be consistent.

 

[Sayantan21]

Can you please tell me your way of solving?

 

[mjc123]

I said ΔH = TΔS at equilibrium, so calculated ΔH at 291K and 1 atm.
ΔH = ΔU + PΔV
Assuming ΔU and ΔV don't change with pressure, d(ΔH)/dP = ΔV
So I calculated ΔH at 100 atm, and then T from T = ΔH/ΔS.
(I could not remember a dP/dT formula off the top of my head.)

 

[Chestermiller]

Since they start out at equilibrium, the Gibbs free energy of the grey start out equal the Gibbs free energy of the white. For small changes in pressure and temperature, we have $$dG=-SdT+VdP$$. So, for the changes in pressure and temperature here, we have for the grey $$dG_g=-S_gdT+V_gdP$$and, for the white,$$dG_w=-S_wdT+V_wdP$$But to maintain equilibrium, these changes must be equal. Therefore, $$\frac{dT}{dP}=\frac{V_w-V_g}{S_w-S_g}=\frac{\Delta V}{\Delta S}$$Of course, specific volume is the reciprocal of density. And, of course, this calculation must be done using consistent units.