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A few lines are written in a book by B.H. Mahan regarding the change in freezing point of water on changing the pressure.

The vapor pressure of water at triple point of water is 4.579 mm. What would happen if the external pressure applied to the system by the piston were to be increased above 4.579 mm ? First of all the va pour present would be completely converted into liquid and solid, to restore the equilibrium pressure.

I'm in total agreement of this. But the next lines appear to be odd.

It is found experimentally that as the pressure on the system is further increased the temperature of the system must be decreased, in order for both ice and water to remain in equilibrium.

I don't understand the second statement. If the equilibrium is further disturbed why the system should not behave the same as before? Why do we have to change the equilibrium constant to restore the equilibrium? Thanks in advance.

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From the triple point (273.16K, 611.657 Pa), if pressure is increased and temperature maintained at 273.16K, all the water being liquid is the equilibrium condition.

To maintain the solid and liquid phases in equilibrium, as pressure is increased above the triple point pressure, temperature must decrease slightly (to 273.15K if the pressure has increased to atmospheric pressure, for example).

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  • $\begingroup$ Can you please explain this from the perspective of equilibrium and le Charlie's principle? $\endgroup$ – user50247 Sep 21 '17 at 16:26
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    $\begingroup$ @gowreeshmago Liquid is the most dense phase of water, so increasing pressure favors the liquid phase with respect to the vapor or solid phase. $\endgroup$ – DavePhD Sep 21 '17 at 16:32

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