Cyclic Processes

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Heat some gas, let it push a piston out, then cool it and push the piston back, until the gas is exactly where it started. Did it do any net work at all? If the answer were always zero, no engine could run, because engines work in cycles. Cycles come up often in JEE, so let's begin.

0:25A cycle ends where it began

So, a cyclic process brings the gas back to its starting state, okay? Same pressure, volume and temperature. Now, internal energy, capital U, depends only on the state; for an ideal gas, only on temperature. So over the full cycle, what is delta U, its total change, written with the little triangle delta? Zero, exactly! The zigzag letter is the Greek capital sigma, meaning the sum over all the steps. So the sum of delta U is zero. Right? Now the first law, in the course's sign convention: capital Q is the heat given to the gas, capital W the work done by it. Add Q equals delta U plus W over all the steps, and the delta U values cancel. The double arrow means so: the sum of Q equals the sum of W, as shown. Clear? Now, a tempting thought. Nothing has changed, so the net heat and the net work must both be zero, right? No, no, careful here, that is the trap! Delta U depends only on the end states, but Q and W depend on the path. So they are equal, but need not be zero. That answers our puzzle. Got it?

1:40Net work is the enclosed area

Now, where is this work on a graph? On the p V diagram, small p, the pressure, goes up, and capital V, the volume, goes across. Work is the integral of p, d V; the tall S is the integral sign, and it means the area under the curve, okay? When the gas expands, its point moves right, and W is greater than zero. When it is compressed, the point moves left, and W is less than zero. With me? In a cycle, the expansion adds its area and the compression takes its area away. What is left is the area enclosed by the loop. Okay? Now the sign. Picture a clockwise loop. Along its top, where the pressure is high, does the gas move right or left? To the right, yes! Clockwise means expanding at high pressure and compressing at low pressure, so the net work is positive. Anticlockwise, it is negative: net work is done on the gas. Right?

2:41The recipe for any cycle

Alright, now the course's recipe for any cycle. Step one: list the states, and find each temperature from p V equals n R T; n is the number of moles, R the gas constant. Okay? Step two: for each step, find W, and delta U from n C v delta T, as shown. C v, with the small v below, is the molar heat capacity at constant volume; this holds in every process of an ideal gas. Clear? Step three: the heat, from the first law, Q equals delta U plus W, as you see. With me? Then check: the delta U values add to zero, and the sum of Q equals the sum of W equals the enclosed area. Got it?

3:27Example 1: heat and work in a rectangular p-V cycle

Now, example one. One mole of a monatomic gas goes round A, B, C, D and back to A, following the arrows. Each point is in brackets as pressure, then volume; the small zero marks the starting values, p naught and V naught. B doubles the pressure, C doubles both, and D doubles only the volume. Find the heat absorbed in each step, and the net work. Pause the video and try it first. Done? Okay, let's check. With n equal to one, T is p V over R, so T A, the T with a small A below, is p naught V naught over R, as shown. Since p V doubles at B and D, and is four times at C, so do the temperatures. With me? From A to B, the volume is fixed, so W is zero. For a monatomic gas C v is three over two R, so Q equals delta U, three over two R times T A: three over two p naught V naught. Okay? From B to C, the pressure is fixed at two p naught and the volume grows by V naught: W is two p naught V naught. Delta U is three over two R times two T A, the dot meaning times: three p naught V naught. So Q is five p naught V naught. Right? Now C to D, then D to A: the temperature falls from four T A to two T A, and then to T A again. So is heat absorbed or given out? Given out, yes! C to D is at fixed volume, so there is no work, and the temperature falls: Q is three over two R times minus two T A, so minus three p naught V naught. Clear? D to A is at fixed pressure p naught, and the volume shrinks by V naught. So W is minus p naught V naught, delta U is minus three over two p naught V naught, and both are negative, so Q is minus five over two p naught V naught. Got it? Now add the heats: three over two, plus five, minus three, minus five over two, which is one. So the net work is p naught V naught. Check: on the p V diagram this is a clockwise rectangle with sides p naught and V naught, so its area is exactly that. Everything matches!

5:49Example 2: V-T rectangle turns anticlockwise on p-V

Now a harder one, example two, on a V T diagram: volume against temperature. Each point of the rectangle is written as temperature, then volume; the ones and twos are small subscripts. T two is more than T one, and V two more than V one. Name the processes, and say whether the cycle is clockwise on the p V diagram. Pause the video and try it first. Done? Okay, let's check. From A to B and C to D, the temperature is fixed: isothermal. From B to C and D to A, the volume is fixed: isochoric, with no work. Easy so far? Now plot T across and V up, as the points are written. The path goes up, right, down, left: clockwise on the V T diagram. Tempting to say it is clockwise on p V too, with positive net work. Is it? No, no, careful here, that is the trap! Work is an area on the p V diagram only, so let's compute it. Per mole, A to B is an isothermal expansion at T one. WAB, with A B small below, is R T one times the natural log, written l n, of V two over V one: positive. Okay? C to D is an isothermal compression at the higher T two. So WCD is R T two times the natural log of V one over V two, or minus R T two times the natural log of V two over V one: negative. With me? Add them: the net work is minus R times, in brackets, T two minus T one, times the natural log of V two over V one. Negative, because more work is done on the gas at the higher temperature than by it at the lower. So the loop is anticlockwise on the p V diagram, not clockwise. Got it?

7:42Quick revision: net Q equals net W, the enclosed area

So, quick revision. One: in a cycle, the sum of delta U is zero, so net heat equals net work. Two: net work is the area enclosed on the p V diagram, positive if clockwise. Three: the recipe: temperatures, then W and delta U, then Q from the first law. Now try the practice problems on this page. All the best!