Hello, dear students, we are studying the chapter, Solution and Colligative Properties, so in this module let me take you through the understanding of vapour pressure versus the temperature, so let us go with that.
Now we will see the distribution of molecular speeds in liquid, as I said that it has similarities to that of gases, so let’s see that first. Well, liquid also has the random motion like gases as well as their distribution is similar to that of velocity distribution of gases. So we have learnt that in earlier modules also. So, and we learnt that this is because of the Maxwell-Boltzmann’s Distribution which is nothing but a plot between a number of molecules versus the kinetic energy. So if we plot that let us see what I can see here, number of molecules versus kinetic energy. Now at a certain temperature T1, this is how the graph varies, this was a similar graph for gases also. Similarly, if at all I increase the temperature, let me take it to T2 temperature, then the graph will vary in this way, that is it will slowly peak will start coming down and it will start becoming more uniform. So in that case now what I could suggest here is let us take a point or kinetic energy after which let’s say liquid particles are starting to escape, right, so this is the kinetic energy, what you can call is the minimum kinetic energy required for molecules to escape from the liquid state. So that is what it is, so it is the minimum kinetic energy. Now if this minimum kinetic energy if at all being drawn towards that curve they will start touching at two different points, then they will generate an area under the curve. So let us see what exactly it represents, so at temperature T1, area under the curve represents the number of particles which are escaping out to the vapour phase. So that yellow coloured is the area which shows that at temperature T1 how many number of particles have escaped. Similarly at temperature T2 you could see this blue coloured shaded part as well as the yellow coloured part, both of them combined are the total number of particles escaping out to the vapour state. So it is very clear that in vapour phase molecules are more at higher temperature, right, as we learnt this earlier also, as the temperature is increased vapour pressure will also increase, right. So what does that mean, you can see evidently here that at temperature T1 the number of molecules escaping are very less but at temperature T2, you could see number of particles escaping are more. So therefore what I could conclude here is low temperature is equivalent to say that it is less likely to vaporize that means it is low vapour pressure. While high temperature it is equivalent to say more likely to vaporize therefore it is at high vapour pressure. So you can now relate how vapour pressure and temperature are co-related to each other, right. So let us try to derive that, now we know that vapour pressure will only be defined when liquid-vapour equilibrium is existing so therefore let me consider that liquid vapour equilibrium is there where K1 is the equilibrium constant and P1 is the vapour pressure. Now therefore K1 is the equilibrium constant at temperature T1 and P1 is the vapour pressure of liquid at temperature at T2.
Now what I can say according to Vant Hoff’s equation we can always relate that as log of K2 by K1 is equal to delta H vaporization by 2.303R into 1 by T1 minus 1 by T2. This is Vant Hoff’s equation where Delta H vaporization is the enthalpy of vaporization, right.
Now observe it carefully, can we write K1 is equal to P1, because yes, liquid does not have any vapour content so therefore vapour phase as P1 so equilibrium constant K1 will be equal to P1. Similarly, if at temperature T2 I can write K2 is equal to P2. So can I change that log of K2 by K1 as P2 by P1. So in that case this Vant Hoff’s equation will be converted to Clausius-Clapeyron’s Equation. So, students, that is how we derive at Clausius-Calpeyron’s equation which is the relation between the vapour pressure of the liquid versus the temperature for a volatile solvent.
So as we go into the next module we will see its applications and more importantly how it will derive towards the Rour’s Law.
Thank you.