# Relative lowering of vapour pressure

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

• Why indeed should the vapour pressure NOT be temperature dependent? Please clarify your question, which one are you unsure about? – Karl Jun 5 '19 at 19:35
• i want mathematical approach to the terms (1 - Ps/Po) that how they balance the effect of temprature! – yaseen wazir Jun 6 '19 at 3:46
• 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? – yaseen wazir Jun 6 '19 at 3:48

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.
• 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. – MasterYoda Jun 6 '19 at 22:04