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Saturation temperature is defined as the temperature at which a pure substance say water, changes phase, at a fixed pressure. So water boils at 100⁰C , if the external pressure is kept constant at 101.3 kPa(1atm). At 12.35 kPa water boils at 50°C, which indicates that there is a pressure-temperature dependence at saturation points, but is this p-t dependence driven by pressure only? i.e can we keep the pressure fixed at 1atm and still boil water at a temperature other than 100°C, by increasing the heat supply and hence increasing temperature.

Also, is this pressure-temperature dependence a consequence of the gibbs-phase rule.

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  • $\begingroup$ The Clausius-Clapeyron equation is the whole story. $\endgroup$ – Bert Barrois Mar 5 '18 at 17:13
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I guess you cannot increase the fluid temperature while keep the pressure constant when it is boiling. The temperature is fixed once the pressure is fixed determined by the fluid vapor pressure curve. Practically, it is more feasible, as you said, to adjust pressure. For example in a high pressure cook, the weight is used to adjust pressure and thus the cooking (boiling) temperature is increased.

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No as long as pressure stays the same, boiling point remains the same. Pressure and temperature are porportional so if you increase pressure you increase temperature, decreasing the amount of energy required to boil the liquid so decreasing boiling point so pressure and boiling point are inversely proportional.

Not sure about the gibbs-phase rule.

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  • $\begingroup$ Can you elaborate on 'pressure and boiling point are inversely proportional'. If i decrease the external pressure as is the case at higher altitudes, the boiling point/boiling temperature also decreases, which is a direct relationship. $\endgroup$ – Mohammad Nayef Mar 5 '18 at 7:06
  • $\begingroup$ I have to specify. For a given volume, pressure and temperature are porportional or increase pressure, increase temp. For a increase in pressure, boiling point decreases because you have increased temp thru pressure decreasing boiling point. That is for a contained volume. $\endgroup$ – Matthew Fizer Mar 5 '18 at 11:41

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