Work on a P–V Diagram

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. We take some gas round a loop. It expands, it is compressed, and it comes back to exactly where it started: same pressure, same volume, same temperature. So overall, it did no work, right? Not so fast! Today you will see such a gas do twelve hundred joules of work, while its internal energy does not change at all. So where does that energy come from? Questions on P V diagrams come up often in JEE, so let's begin.

0:35Work is the area under the path

So, look at these three graphs. Pressure P goes up, and volume V goes across. In each, the gas expands at constant temperature from V one to V two; the small one and two just mark the start and the end. Graph a: in one step, against the final pressure, P two. Graph b: in two steps. Graph c: along the smooth reversible curve, infinitely many tiny steps. Okay? Now, the shaded part under each path. That area is the work done by the gas. This is the key idea: work done by the gas equals the area under its path. Three paths, three different areas: for the same start and end, the least work in one step, and the maximum along the reversible curve, n R T times the natural log of V two over V one, from our lesson on reversible isothermal processes. So the work depends on the path. Clear? Now the sign. In our sign rule, small w is the work done on the gas. So the work done by the gas, w with the subscript by, is minus w. Expansion gives positive work by the gas, the plus sign; compression gives negative, the minus sign. Right? And one useful conversion. One litre, capital L, is ten to the power minus three cubic metres; one bar is ten to the power five pascals, P a. Multiply: one litre bar is one hundred joules, capital J for joules. We will use this in both examples. Got it?

2:08Two special paths: vertical and straight lines

Now, two special paths. First, isochoric, which means constant volume. What does that path look like, and how much area is under it? Yes, a vertical line! The pressure changes but the volume stays fixed, so the path goes straight up or straight down. Now, a line has no width: no area, and no work. With me? Second, a straight line, from P one, V one in brackets, to P two, V two. Under it, we get two vertical sides and a slanting top: a trapezium. Okay? Area of a trapezium: one half, times the sum of the parallel sides, times the width. So w by equals one half, times P one plus P two, times V two minus V one. Simply, the average pressure times the change in volume. Clear?

2:57Round a cycle: the enclosed area

Now the big idea: cycles. In a cycle, the gas goes round a closed loop and comes back to its starting state, just like in our puzzle. Right? Picture a loop. The gas goes out to the right along the top, and comes back along the bottom. Going out, it expands: positive work, the area under the top. Coming back, it is compressed: negative work, the area under the bottom. So what is left over? The area between them, yes! The net work done by the gas is the area enclosed by the loop. Okay? Now, if the loop runs clockwise, is the net work done by the gas positive or negative? Positive, correct! Clockwise, the gas expands along the top at high pressure, and is compressed along the bottom at low pressure, so it does more work than is done on it. Anticlockwise is the opposite: negative, so net work is done on the gas. Got it? Now, the gas is back at its starting state, so delta U, the change in internal energy, is zero. The Greek letter delta, the little triangle, means change in. So the heat, small q, is zero too, right? No, no, careful! That is the trap. By the first law, delta U equals q minus w by, and that is zero. The double arrow means which gives: q equals w by. Net heat absorbed equals net work done by the gas. And that answers our puzzle: the energy for the work comes in as heat. Clear?

4:29Topper's trick: three quick steps

Now, a topper's trick. One: spot the shape on the diagram, a rectangle, a triangle or a trapezium. Okay? Two: find its area in litre bar, then times one hundred for joules. With me? Three: take the sign from the direction. Clockwise or expansion, plus; anticlockwise or compression, minus. That's it. Let's try it.

4:52Example 1: clockwise rectangle cycle, work and heat

Okay, Example one. Some gas is taken clockwise round a rectangle, between one and four bar, and between two and six litres. Find the net work done by the gas, and the net heat absorbed. Pause the video, and try it first. Done? Okay, let's check. The shape is a rectangle: four minus one, three bar tall, and six minus two, four litres wide. Right? The cycle runs clockwise. So, is the net work done by the gas positive or negative? Positive, very good! The area is three times four, twelve litre bar, and one litre bar is one hundred joules: so, twelve hundred joules. With me? Now, the gas came back to its starting state. How much heat did it absorb? Twelve hundred joules, exactly! For a cycle, delta U is zero, so the heat absorbed equals the work done by the gas. There is our puzzle: the same state at the end, twelve hundred joules of work, paid for by twelve hundred joules of heat. Mystery solved!

6:00Example 2: straight-line path, work, delta U and q

Now Example two, a little harder. We take a monatomic ideal gas along a straight line, from two bar and one litre, to one bar and three litres. Find the work done by it, delta U, and q. Pause and try. Hint: a trapezium. Done? Okay, let's check. The volume goes from one litre to three, so delta V, the change in volume, is two litres. Now, which pressure do we multiply it by: two bar, or one bar? Neither! The pressure changes all along the path. The area is a trapezium, so use the average pressure: one half, times two plus one, times three minus one. Three litre bar, that is three hundred joules, positive because the gas expands. Okay? Now, what about delta U? The pressure fell, from two bar to one bar. So the gas cooled down, right? Oh no, careful! Look at P times V, not just the pressure. For a monatomic ideal gas, delta U is three over two n R delta T, with n the moles, R the gas constant and T the temperature. Since P V equals n R T, delta U is three over two, times P two V two minus P one V one. With me? Now the numbers. P two V two is one times three, three; P one V one is two times one, two. So P V went up, which means the temperature went up! Delta U is three over two times one: one point five litre bar, or one hundred and fifty joules. Right? Last, the heat. Delta U equals q minus w by, so q equals delta U plus w by: one hundred and fifty plus three hundred, four hundred and fifty joules absorbed. That's it!

7:53Quick revision: area under path, enclosed area, signs

So, quick revision. One: the work done by the gas is the area under its path; an isochoric, vertical line gives none, and a straight line gives a trapezium. Two: round a cycle, the net work is the enclosed area, plus if clockwise, minus if anticlockwise. Three: for any cycle, delta U is zero, so the net heat absorbed equals the net work done by the gas. Now go and try the practice problems on this page. All the best!