The densities of liquid and solid bismuth are 10.0 g/c.c. (10x10 kg/m ) and 9.673 g/c.c. (9.673 x IO3 kg/m3) respectively at the normal melting point 270°C (543 K). The heat of fusion is 2.633 kcal/mole (11.02 kJ/mol). Calculate the melting point of bismuth under a pressure of 100 atm (101,325 x 22 10 N/m ) . Atomic weight of bismuth is 209.

To solve this problem, we can use the Clausius-Clapeyron equation, which describes the phase transition between two states of matter (in this case, solid and liquid). The equation is as follows:

ΔT/ΔP = TΔV/ΔH

where:

• ΔT is the change in temperature (which we are trying to find),
• ΔP is the change in pressure (which is 100 atm or 10132500 Pa),
• T is the initial temperature (which is 543 K),
• ΔV is the change in volume (which we can calculate from the given densities and atomic weight of bismuth),
• ΔH is the heat of fusion (which is 11.02 kJ/mol).

First, we need to calculate ΔV. The volume of one mole of bismuth in the liquid state is (209 g/mole) / (10 g/cm³) = 20.9 cm³/mole. In the solid state, it is (209 g/mole) / (9.673 g/cm³) = 21.6 cm³/mole. Therefore, ΔV = 21.6 cm³/mole - 20.9 cm³/mole = 0.7 cm³/mole = 0.7 x 10^-6 m³/mole.

Now we can substitute all the known values into the Clausius-Clapeyron equation:

ΔT/10132500 Pa = 543 K * 0.7 x 10^-6 m³/mole / 11.02 kJ/mole ΔT = 10132500 Pa * 543 K * 0.7 x 10^-6 m³/mole / 11.02 kJ/mole ΔT = 0.35 K

Therefore, the melting point of bismuth under a pressure of 100 atm is 543 K + 0.35 K = 543.35 K.