Question

2) The temperature dependence of the vapor pressure of solid and liquid mercury can be represented by: in Park = 23.912_8781K in Poud –16.85370$IK tor where T is the Kelvin temperature. Use the Clapeyron equation to calculate the enthalpies of sublimation and vaporization of mercury. In addition, use the data given to calculate Mercury's triple point.

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Answer 1

Using Clausius Clapeyron equation:

d(ln P) / dT = $\Delta$ H_Sublimation / RT² (1)

Consider, d(ln P) / [d (1/T)]

Multiplying both numerator and denominator by dT:  

[d(ln P) / dT] X [ dT / d(1/T)] = T² d(ln P) / dT

Using (1) : T² d(ln P) / dT =  $\Delta$ H_Sublimation / R

In this case,   $\Delta$ H_Sublimation / R = 8781 $\Rightarrow$$\Delta$ H_Sublimation = 8.314 J/mol X 8781 = 73005.234 J/mol

Similary,  $\Delta$ H_vaporization / R = 7051 $\Rightarrow$$\Delta$ H_vaporization = 8.314 J/mol X 7051 = 58622.014 J/mol

Let  T_TP = triple point temperature and  P_TP = Triple point pressure

At the triple point, the vapor pressures of the solid and liquid are equal.

So, 23.912 - 8781 K/T_TP = 16.853 -7051 K/T_TP

7.059 = 1730 K/T_TP    $\Rightarrow$   T_TP = (1730/7.059) K

T_TP = 245.077 K

Now, ln (P_TP /torr) = 23.912 - (8781 / 245.077) = 23.912 - 35.83 = - 11.918

P_TP = 6.669 X 10^-6 torr

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