Why is lowering of vapour pressure (Po - Ps) temperature dependent? And why is relative lowering of vapour pressure (1 - Ps/Po) independant of temperature?

  • $\begingroup$ Why indeed should the vapour pressure NOT be temperature dependent? Please clarify your question, which one are you unsure about? $\endgroup$
    – Karl
    Commented Jun 5, 2019 at 19:35
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    $\begingroup$ i want mathematical approach to the terms (1 - Ps/Po) that how they balance the effect of temprature! $\endgroup$ Commented Jun 6, 2019 at 3:46
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    $\begingroup$ also in case of (Po - Ps) why it depends on temp? because when temprature is increased both Po and Ps increase, they should have balance each other? why they don't balance? $\endgroup$ Commented Jun 6, 2019 at 3:48

1 Answer 1


In the simplest terms and most convenient definitions, the lowering of vapor pressure is not a colligative property while the relative lowering of vapor pressure is indeed a colligative property.

A bit more explanation: A colligative property is dependent on the moles of solute only. The ratio in the definition you provided eliminates the temperature dependence of formulation (assuming ideal solutions) by making the measurement relative to another quantity, which in this case is the vapor pressure of the pure solution. Otherwise, the influences on vapor pressure are multifactorial. I.e. external pressure may affect it, as well as temperature.

Here is a somewhat hand-wavy out-of-the-pocket calculation to demonstrate. Assume the vapor pressure can be expressed as an ideal gas. Therefore pressure may be expressed using the ideal gas law, $P = \frac{nRT}{V}$. Also know Raoult's law for vapor pressure, $P_s = \chi_s P_o$ where $\chi_s$ is the mole fraction. Using your definitions, we have for absolute pressure difference $$P_o - P_s = P_o - \chi_s P_o = \left(1-\chi_s\right)P_o = \left(1-\chi_s\right)\frac{n_oRT}{V}$$ This is therefore dependent upon temperature. Now do the same for the relative difference in pressures $$1-\frac{P_s}{P_o} = 1 - \frac{\chi_sP_o}{P_o} = 1-\chi_s$$ The expression is now dependent upon the moles only, i.e. a colligative property. Of course, in reality this depends on the assumptions we made under ideal Raoult conditions.

  • $\begingroup$ is volume same for the term Po and Ps? or if it is different, can we mention Vo and Vs instead of just V? $\endgroup$ Commented Jun 6, 2019 at 3:44
  • $\begingroup$ I wrote this under the assumption that volume does not change. You can create this isochoric process experimentally when adding solute if needed. But for this particular problem, it does not matter if volume is constant and you can say it changes from $V_o$ to $V_s$. I just wanted to explicitly show how the absolute pressure difference can be a function of temperature. For instance, we can leave the math at$$P_o-P_s=(1-\chi_s)P_o$$and make that argument that because the pressure $P_o$ is present, then the difference in pressure is still subject to all the same forces that change it, like temp. $\endgroup$
    – MasterYoda
    Commented Jun 6, 2019 at 22:04
  • $\begingroup$ I don't understand how lowering of vapour isn't a colligative property. Based on the equation you provided for lowering of vapour pressure, it is dependent on T, V and n sub o. I don't see any variable that accounts for the nature of solute. Is it necessary for a colligative properties to be ONLY dependent on number moles? If it depends on temperature but its still independent of nature of solute then it's not colligative property? $\endgroup$
    – Spluesh
    Commented Jan 7 at 11:11

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