Question

( P4B.7 The enthalpy of fusion of mercury is 2.292 kJ mol' at its normal freezing point of 234.3 K; the change in molar volume on melting is +0.517 cm'mol. At what temperature will the bottom of a column of mercury (mass density 15.6g cm) of height 10.0 m be expected to freeze? The pressure at a depth a in a fluid with mass density pis pgd, where g is the acceleration of free la 9.81 m s?

Answer 1

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From Claperyon equation, we have,

P2 - P1 = ffus. In

We know that, Pressure at the bottom of the mercury column will be the sum of atmospheric pressure and pressure of Hg in the column.

::P2 = (p.g.d) + (Atm pressure)

:: P1 = atmospheric temperature

Therefore, Change in pressure is given by,

$\Delta P=P_{2}-P_{1}=\rho.g.d$

We have, d= 10 m

Therefore, Change in pressure is given by,

$\Delta P=\rho.g.d=(13600\;kg/m^{3})\times9.8\;m/s^{2}\times10\;m$

1 g/cm³= 1000 kg/m³.

$\Delta P=1332800\;Pa$

Given that, Enthalpy of fusion of mercury = 2.292 kJ/mol.

Normal freezing point = 234.3 K

Therefore, Freezing point at the bottom is given by:

$\Delta P=\frac{\Delta H_{fus}}{\Delta V}.\ln\frac{T_{2}}{T_{1}}$

$1332800\;Pa=\frac{2292 \;J/mol}{0.517\times10^{-6}\;m^{3}/mol}.\ln\frac{T_{2}}{234.3\;K}$

1 cm³ =10^-6 m³

1 kJ =1000 J

$\therefore \ln\frac{T_{2}}{T_{1}}=\frac{1332800\;Pa\times (0.517\times10^{-6}\;m^{3}/mol)}{2292 \;J/mol}$

$\therefore \ln\frac{T_{2}}{T_{1}}=3.006\times10^{-4}$

$\therefore T_{2}=234.3\;K\times1.000300681$

$\boldsymbol{\therefore T_{2}=234.370\;K}$

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