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
