Question

Gold has an equilibrium melting temperature of 1337 K at 1 atm and an enthalpy of fusion of 12,550 J/mol. Find the pressure at which the equilibrium melting temperature is 1400 K.
Using this answer, calculate the entropy change for melting 1 mol of gold. The densities of solid gold and liquid gold are 19.3 g/mL and 17.3 g/mL and are assumed to be approximately constant under all conditions

Answer 1

1) Using Clausius Clapeyron equation,

ln(P/P₀) = (H/R) {( 1/T₀)-(1/T)}

ln (P/1 atm) = (12550 / 0.0821) {(1/1337) - (1/1400)}

   ln (P/1 atm) = 152862.363 (0.00075 - 0.0007143)

   ln (P/1 atm) = 152862.363 * 0.0000357

  ln (P/1 atm) = 5.46

  P/1 atm = 10^-5.46

     P/1 atm = 0.0000035

P = 0.0000035 atm

2) Using Gibbs Duhem equation,

S = H / T

= 12550 J/mol / (1400- 1337)K

S = 199.21 J/mol.K