The solid-state transition of Sn (gray) to Sn (white) is in equilibrium at 18.0 ˚C and 1.00 atm, with an entropy change of 8.8 J K-1 mol-1. The density of gray tin is 5.75 g cm-3; that of white tin is 7.28 g cm-3. Calculate the transition temperature at a pressure of 100.0 atm.

Transition temperature = 13 C

Explanation:

ΔS (transition) = 8.8 J/K. mol

ρ (gray) = 5.75 g/cm³ = 5750 kg/m³

ρ (white) = 7.28 g/cm³ = 7280 kg/m³

ΔP = 100 atm = 100 x 101325 = 10132500 Pa

M (Sn) = 118.71g/mol = 118.71 x 10⁻³ kg/mol

T (i) = 18 C, T (f) = ?

We know that

G = H - TS, where G is the gibbs free energy, H is the enthalpy, T is the temperature and S is the entropy

H = V (m) x P, hence the equation becomes

G = V (m) x P - TS

The change in Gibbs free energy going from G (gray) to G (white) = 0 as no change of state takes place hence it can be said that

ΔG (gray) - ΔG (white) = 0

replacing the G with it formula shown above we can arrange the equation such as

0 = V (m) (gray) - V (m) (white) x ΔP - (ΔS (gray) - ΔS (white)) x ΔT

solving for ΔT we get

ΔT = {V (m) (gray) - V (m) (white) x ΔP} / (ΔS (gray) - ΔS (white))

ΔT = {M (Sn) (1/ρ (gray) - 1/ρ (white) x ΔP} / (ΔS (gray) - ΔS (white))

ΔT = {118.71 x 10⁻³ x { (1/5750) - (1/7280) } x 10132500} / (8.8)) = 5.0 C

ΔT = T (initial) - T (transition)

T (transition) = T (initial) - ΔT = 18 - 5 = 13 C