 The photo you see here is of Joseph Louis Gelussac and in early 1800s Gelussac was experimenting with how different volumes of gases combine and based on his calculations another chemist arrived at a relationship involving gases. So this relationship which is called Gelussac's law or the pressure law can be stated as pressure is directly proportional to the temperature but this relationship between the pressure and the temperature of the gas assumes two conditions. One that the volume occupied by the gas is fixed and two that the amount of gas is also fixed. So let me just write down both of these conditions here. To understand this better we can consider this setup where we have this box and in this box we have a few molecules of the gas and because our box is a rigid box and the gas is within this box the volume is now fixed and also because we are not adding more gas or removing the gas from this box the amount of gas in this case is fixed. So now both of these conditions are met and we know that these gas molecules will be randomly moving around and while they are randomly moving about like this they could be colliding with any of these phases of the box. So now to better understand this relationship let's take another box and this box is an identical box with the same amount of gas in this box as well and in this box also the gas molecules will be moving in random directions. So now that we have two identical boxes and for both of them both of these conditions are met. What if we heat this box when we are heating this lower phase of the box what happens is because of the heat the kinetic energy of these molecules increases which means that their velocities also increase and they start moving faster in the box and we know that when the kinetic energy increases the momentum will also increase and we define force as the rate of change of momentum. So effectively what's happening is when we heat this phase as the molecules collide with this phase and their kinetic energy increases they collide with each of these phases faster and because of this increased momentum the force applied on each of these phases increases and we know that the pressure is defined as force upon area. So since the force increases the pressure also increases. So what happened here was by increasing the temperature we see how the pressure also increases but this was a very qualitative description of what's happening. Let's see how we can express this mathematically and what new information we can get from there. So I have written down our earlier relationship between pressure and temperature like this that is P is proportional to T where the P is pressure and T is temperature. I can get rid of the proportionality sign and write it in the form of an equation as P is equal to some M times T where M is a constant and here you can see that M is a dimensional constant. So going back to our molecules in a box if we know the temperature at which this box is kept and if we know the pressure at that point we can get the value of this M and it will be in some pressure per temperature units. So now if we just rearrange this we can write this as P divided by T is constant and what this rearrangement does is that it simplifies some calculation for us. So again if we go back to this box let's say we started at some temperature T1 and at that time the pressure was P1 and we then increase the temperature to T2 because of which the pressure then became P2. We know that the ratio P by T is constant. So we can write this as P1 by T1 is equal to P2 by T2 because we know that the ratios of pressure and temperature will be constant. If we know any three of these quantities we can easily find the fourth one. Now again if we look at this equation we can see that this resembles the equation of a straight line. If you think of the equation of a straight line written as y is equal to ax plus b and let's say if this intercept is 0. So we can write the equation of the line as something like y is equal to ax and this equation here resembles this form. So if we were to graphically represent this relationship if we have pressure on the y-axis and temperature on the x-axis we can show this relationship as a straight line but there is one very important point to note here. This temperature that we are talking about here is in the form of Kelvin and not in the form of degree Celsius and not just here whenever we write this relationship in this form or in the form of a ratio everywhere the temperatures that we take are in Kelvin and so if the temperatures which are known if they're given in degree Celsius what we do is we add 273 to it and we can get the temperatures in Kelvin which we can use in these equations. So now let's go through one example problem and see how these calculations are done. Now let's look at this problem which is asking us to calculate the pressure change when a constant volume of gas at a pressure of 180 is heated from 20 degree Celsius to 30 degree Celsius. So if you want you can pause the video here and give this a try. Okay so let's write down what is known from the information given to us. So first we can see that we've been told the gas has a constant volume and so if it is in a box like we saw before the amount of gas is also fixed. So in that case we can apply Galer-Sachs law and we can write P1 by T1 will be equal to P2 by T2 where we know that our initial pressure is 180 and initial temperature is 20 degree Celsius. We know that the final temperature is 30 degree Celsius. So if we rearrange this we can write P2 as T2 divided by T1 into P1 and as we saw before the temperatures that we use here should be in Kelvin and not in degree Celsius. So if we directly substitute 30 and 20 here instead of T2 and T1 we'll get the wrong answer. So to use this relation we will first convert these temperatures into Kelvin. So we will add 273 to both of these. So now the initial temperature becomes 293 Kelvin and the final temperature becomes 303 Kelvin and so we can substitute these two temperatures and the initial pressure in this equation to calculate P2 which is 303 divided by 293 into 1.08 m and if you solve this we get P2 to be 1.03 8 m and then to calculate the pressure change what we're going to do is the pressure change will be equal to P2 minus P1 which is this 1.03 minus the initial pressure which is 180 m and we get the pressure change to be 0.03 8 m.