Part 3, please explain Based on the following data of a pure substance (with mol

Question

Part 3, please explain

Based on the following data of a pure substance (with molar mass 160 g//mol), determine its heat of vaporization, heat of sublimation, and heat of fusion by assuming that they are independent of temperature and pressure. (Note that 1 bar = 100 k Pa and 1 atm = 1.01325 bar) Triple point: -8.0 degree C and 6.0 k Pa Normal freezing point (at 1 atm): -7.0 degree C Normal boiling point (at 1 atm): 58 degree C Density of liquid: 3.119 g/mL Density of solid: 4.05 g/mL Plot the phase diagram (pressure in k Pa vs. temperature in degree C) to cover the trip point, the normal freezing point, the normal boiling point, and the three phase boundaries of this pure substance. Determine the temperature at which the equilibrium vapor pressure of the substance is equal to 0.010 bar. Is the substance in its liquid state or solid state at this condition?

Explanation / Answer

Apply Clasius Clapeyron equation

ln(P2/P1) = H/R*(1/T1-1/T2)

ln(0.01/1) = H/8.314*(1/(58+273) - 1/T2)

We need Hvap

so.. from previous data

ln(P2/P1) = H/R*(1/T1-1/T2)

choose triple point + normal boiling point:

T1 = 58°C = 58+273 = 331 P1 = 1 bar

T2 = -8°C = -8+273 = 265 P2 = 6 kPa = 0.06 bar

ln(0.06/1) = H/8.314*(1/(331) - 1/265 )

H = ln(0.06/1) *8.314 / (1/(331) - 1/265 )

H = 31086.59 J/mol

now, use in our eqution

ln(0.01/1) = H/8.314*(1/(58+273) - 1/T2)

ln(0.01/1) = 31086.59 /8.314*(1/(58+273) - 1/T2)

solve for T2

ln(0.01/1) /31086.59 *8.314 - (1/(58+273) = - 1/T2)

-0.00425 = -1/T2

T2 = 0.00425^-1 = 235.294 K = -37.706 °C

this s going to be liquid/vapor phase

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