Respuesta :
Explanation:
Relation between pressure, latent heat of fusion, and change in volume is as follows.
[tex]\frac{dP}{dT} = \frac{L}{T \times \Delta V}[/tex]
Also, [tex]\frac{L}{T} = \Delta S^{fusion}_{m}[/tex]
where, [tex]\Delta V^{fusion}_{m}[/tex] is the difference in specific volumes.
Hence, [tex]\frac{dP}{dT} = \frac{\Delta S^{fusion}_{m}}{\Delta V^{fusion}_{m}}[/tex]
As, [tex]\Delta S^{fusion}_{m} = \frac{L}{T} = \frac{6010}{273.15}[/tex] = 22.0 J/mol K
And, [tex]\Delta V^{fusion}_{m} = \frac{M}{d_{H_{2}O}} - \frac{M}{d_{ice}}[/tex] ...... (1)
where, [tex]d_{H_{2}O}[/tex] = density of water
[tex]d_{ice}[/tex] = density of ice
M = molar mass of water = [tex]18.02 \times 10^{-3} kg[/tex]
Therefore, using formula in equation (1) we will calculate the volume of fusion as follows.
[tex]\Delta V^{fusion}_{m} = \frac{M}{d_{H_{2}O}} - \frac{M}{d_{ice}}[/tex]
= [tex]\frac{18.02 \times 10^{-3}}{997} - \frac{18.02 \times 10^{-3}}{920}[/tex]
= [tex]-1.51 \times 10^{-6}[/tex]
Therefore, calculate the required pressure as follows.
[tex]\frac{dP}{dT} = \frac{22}{-1.51 \times 10^{-6}}[/tex]
= [tex]1.45 \times 10^{7} Pa/K[/tex]
or, = 145 bar/K
Hence, for change of 1 degree pressure the decrease is 145 bar and for 4.7 degree change dP = [tex]145 \times 4.7 bar[/tex]
= 681.5 bar
Thus, we can conclude that pressure should be increased by 681.5 bar to cause 4.7 degree change in melting point.
