Real Gases

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Squeeze one mole of carbon dioxide into a one litre bottle at three hundred kelvin. What pressure does it exert? Just use p V equals n R T, right? But here is the surprise. The van der Waals equation, built for real gases, gives about ten percent less. So where did the missing pressure go? Let's find out.

0:26When is a real gas nearly ideal?

So, start with the law you know, p V equals n R T, okay? Small p is the pressure, capital V the volume, and n the number of moles. Capital R is the gas constant, and capital T the temperature in kelvin. Right? This law is for an ideal gas. In the ideal model, the molecules have negligible size, and there are no forces between them except during collisions. But real molecules do have a size, and they do attract each other. Clear? So when does a real gas behave almost like an ideal gas? At high or low pressure? And at high or low temperature? Low pressure and high temperature, exactly! So real gases deviate the most at the other end, at high pressure and low temperature. Okay? Now let me ask you why. At a very high pressure, the gas is squeezed into a small space. What about the volume of the molecules themselves? Yes, that is the point! Their own volume becomes a larger fraction of the container, so we can no longer call their size negligible. Got it? And at low temperature, the molecules have less kinetic energy, the energy of motion, so the attractive forces matter more compared with it. Both ideal assumptions fail together. With me?

1:49The van der Waals equation

Now, the fix: the van der Waals equation. In the first big round bracket, we have p plus small a times n squared, over V squared. That multiplies the second bracket, V minus n times small b, and the product equals n R T, okay? Small a and small b are the van der Waals constants; each gas has its own. Right? Small b corrects for the finite size of the molecules, so the volume V becomes V minus n times b. Clear? And small a corrects for their attraction, so the pressure p becomes p plus small a times n squared, over V squared. Okay? One more fact from the course. The wavy equals sign means approximately. So small b is not the volume of the molecules itself; it is about four times their actual volume in one mole. Got it? Now, a quick exercise. Take one mole, so n equals one, and solve this equation for p. Try it in your head. So, did you get this? p equals R T over, V minus b, minus small a over V squared. Now look at the sign of the small a term. Inside the bracket it had a plus, but once we move it to the other side, it becomes a minus. Many students keep the plus here and lose the marks, so be careful. With me?

3:10The compressibility factor Z

Now, to measure how far a gas is from ideal, we use the compressibility factor, capital Z, which equals p V, over n R T, okay? So tell me, for an ideal gas, what is the value of Z? It is one, exactly! For an ideal gas, p V equals n R T, so the top and the bottom are equal. So a value of Z different from one means the gas is not behaving ideally.

3:39Example 1: Boyle temperature of nitrogen from a and b

Clear? Now, example one. For nitrogen, small a is zero point one three seven pascal metre to the power six per mole squared, and small b is three point eight seven times ten to the power minus five cubic metres per mole. Estimate the Boyle temperature, T with a small subscript B, equal to small a, over R times small b. At this temperature the gas obeys Boyle's law over a wide range of pressure. Pause the video and try it first. Okay, let's check. First, which value of the gas constant R do we use here? Look at the units of small a and small b: pascals, metres and moles, all S I units. So R is eight point three one four joules per mole per kelvin. Right? Now put in the numbers: zero point one three seven, over eight point three one four times three point eight seven times ten to the power minus five. Okay? The bottom is about three point two one seven five times ten to the power minus four. Do not drop that power of ten! Clear? And zero point one three seven divided by that gives about four hundred and twenty six. So the Boyle temperature of nitrogen is about four hundred and twenty six kelvin. Simple, isn't it?

5:02Example 2: van der Waals vs ideal pressure of CO2 in 1 L

Now, example two, and this one solves our puzzle. One mole of carbon dioxide, C O two, with the two written as a small subscript, occupies one point zero zero litres at three hundred kelvin. Here small a is zero point three six four pascal metre to the power six per mole squared, and small b is four point two seven times ten to the power minus five cubic metres per mole. Compare the van der Waals pressure with the ideal pressure. Pause the video and try it first. Okay, let's check. The volume is one point zero zero litres. So can we put one point zero zero straight in for the volume? No, we cannot, so be careful here! Our small a and small b are in S I units, so the volume must be in cubic metres. One litre is ten to the power minus three cubic metres. Right? Now the first term. R T is eight point three one four times three hundred, about two thousand four hundred and ninety four, and V minus b is nine point five seven three times ten to the power minus four. Dividing gives about two point six zero five times ten to the power six pascals. Okay? Next, the attraction term, small a over V squared. Zero point three six four, divided by ten to the power minus six, gives zero point three six four times ten to the power six pascals. Got it? Now we subtract: two point six zero five minus zero point three six four, all times ten to the power six. So the van der Waals pressure is about two point two four times ten to the power six pascals. With me? For the ideal gas, p is just R T over V, two thousand four hundred and ninety four divided by ten to the power minus three. That gives about two point four nine times ten to the power six pascals. Clear? So the van der Waals pressure is about ten percent lower. There is our missing pressure from the start! Now, which effect won, the size of the molecules or their attraction? The attraction, yes! The size term alone pushed the pressure above the ideal value, but the attraction term pulled it down by more. Right? The course takes this one step further. With n equal to one, Z equals p V over R T: two point two four times ten to the power six, times ten to the power minus three, divided by two thousand four hundred and ninety four. That gives about zero point nine zero. It is less than one, which tells us that attraction dominates. Well done!

7:50Quick revision: when real gases are nearly ideal, a, b and Z

So, quick revision. Real gases are nearly ideal at low pressure and high temperature. The van der Waals equation corrects the ideal law, with small a for attraction and small b for size. And the compressibility factor Z equals one for an ideal gas. Now go and try the practice problem on this page. All the best!