Question

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.

Answer 1

Answer:

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