Question

The enthalpy of fusion of mercury is 2.292 kJ mol-1 at its normal freezing point of 234.3 K; the change in molar volume on melting is +0.517 cm3 mol-1. At what temperature will the bottom of a column of mercury (mass density 13.6 g cm-3) of height 10 m be expected freeze? The pressure at a depth d in a fluid with mass density rho is rhogd, where g is the acceleration of free fall, 9.81 ms-2.

Answer 1

Answer:

T_2= 234.37 K

Explanation:

According to Claperyon, we know that

$P_2-P_1= \frac{\Delta H_{fus}}{\Delta V}\times\frac{T_1}{T_2}$

P_1= Atmospheric pressure 760 mm Hg

P_2 = pressure at the bottom of the column

= 10×10^3 mm of Hg+ 760 mm of Hg

= 10760 mm of Hg

now,

P_2-P_1= 10760-760= 10^4 mm

P_2-P_1 ( in pascals) = 10^4× 133.322= 1333220 mm

the enthalpy of fusion (ΔH-fus) of mercury is 2.292 KJ/mol

use the above equation to calculate ΔT as follows

$1333220= \frac{2292}{0.517\times10^{-6}}\times ln\frac{T_2}{234.3K}$

therefore, T_2= 234.37 K