where is the molar mass. The densities of ice and liquid water are 0.917 g M and 1.000 g dm-3 and the molar mass, = 18.02 g, so that the molar volumes of M ice and liquid water are = 18.02 g / 0.917 g dm-3 = 19.58 dm3 Vice and = 18.02 g / 1.000 g dm-3 = 18.02 cm3 Vwater respectively. The volume change on melting is therefore = - =18.02 - 19.58 = -1.56 cm3. Vwater Vice ΔfusV We must remember to convert this value into units of m3 = -1.56 cm3 = -1.56 × 10-6 m3 ΔfusV before substituting it into the Clapeyron equation. We already know one point on the solid-liquid equilibrium line since we know that ice melts at a temperature of = 273.15 K and pressure of = 101325 Pa. The T1 p1 enthalpy of fusion of ice = 6.030 kJ mol-1. ΔfusHo Thus at a pressure of 70 bar ( 7093000 Pa, which is typical of the pressure exerted by an ice skater and so Thus at a pressure of 70 bar, the melting point of ice is = 272.66 K, a decrease of T2 0.49 K. The melting temperature of ice is therefore lowered by the effect of increased pressure. The application of pressure to a block of ice held at a constant temperature may therefore cause melting. This is unusual; for most substances, the melting point increases with pressure. The peculiar behaviour of water arises because the density of liquid water is greater than that of ice. The contraction on melting causes results in a negative value for and so the solid-liquid line on the pressure-temperature ΔfusV
where M is the molar mass. The densities of ice and liquid water are 0.917 g and 1.000 g dm-3 and the molar mass, M = 18.02 g, so that the molar volumes of ice and liquid water are Vice = 18.02 g / 0.917 g dm-3 = 19.58 dm3 and Vwater = 18.02 g / 1.000 g dm-3 = 18.02 cm3 respectively. The volume change on melting is therefore ΔfusV = Vwater - Vice =18.02 - 19.58 = -1.56 cm3. We must remember to convert this value into units of m3 ΔfusV = -1.56 cm3 = -1.56 × 10-6 m3 before substituting it into the Clapeyron equation. We already know one point on the solid-liquid equilibrium line since we know that ice melts at a temperature of T1 = 273.15 K and pressure of p1 = 101325 Pa. The enthalpy of fusion of ice ΔfusHo = 6.030 kJ mol-1. Thus at a pressure of 70 bar ( 7093000 Pa, which is typical of the pressure exerted by an ice skater and so Thus at a pressure of 70 bar, the melting point of ice is T2 = 272.66 K, a decrease of 0.49 K. The melting temperature of ice is therefore lowered by the effect of increased pressure. The application of pressure to a block of ice held at a constant temperature may therefore cause melting. This is unusual; for most substances, the melting point increases with pressure. The peculiar behaviour of water arises because the density of liquid water is greater than that of ice. The contraction on melting causes results in a negative value for ΔfusV and so the solid-liquid line on the pressure-temperature