Reversible Isothermal Processes

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Gas in a cylinder pushes a piston out, and does over seven kilojoules of work. But its temperature stays exactly the same. For an ideal gas, that means its internal energy does not change at all. So where did those seven kilojoules come from? Energy from nowhere? Hmm. Hold that thought. This topic comes up often in JEE, so let's begin.

0:28What makes a process reversible?

So, what does reversible really mean? Take some gas in a cylinder, held in by a piston. Suppose I suddenly drop the outside pressure by a lot. What does the gas do? It rushes out, exactly! Fast, uneven, nowhere near equilibrium. That is not reversible. In a reversible process, the gas passes through a series of equilibrium states. Okay? The way to do it: keep the external pressure, the P with the little subscript e x t, within an infinitesimal amount of the gas pressure, a tiny, tiny bit. Then the gas is practically in equilibrium at every moment. With me? Of course, no real piston can do this perfectly: it would need infinitely many tiny steps, and so an infinite time. So a truly reversible process is an idealisation. Right?

1:18The work in a reversible isothermal change

Now, the main result. Isothermal means the temperature, capital T, stays constant. For n moles of an ideal gas, w rev equals minus n R T, times the natural log of V two over V one. Small w with the subscript r e v is the reversible work, R is the gas constant, l n is the natural log, and V one and V two are the initial and final volumes. Okay? Now you tell me. In an expansion the gas grows, so V two is more than V one. Is w positive or negative? Negative, correct! On screen: V two greater than V one, and the double arrow means which gives, w less than zero. The ratio is above one, so its log is positive, and the minus sign makes w negative. Got it? Compression is the opposite: V two is less than V one, the log is negative, and w is positive. Work is done on the gas. Put in final over initial, and the sign takes care of itself. Clear?

2:19Log to base ten, and the pressure form

Now, the same formula with the log to base ten. w rev equals minus two point three zero three, n R T, log of V two over V one. The two point three zero three converts the natural log to base ten. Don't forget it! Okay? Now, sometimes the question gives you pressures, not volumes. Same pattern, final over initial, so it becomes log of P two over P one, right? No, no, careful! That is the trap. At constant temperature, for an ideal gas, P one V one equals P two V two. So V two over V one equals P one over P two. The ratio flips! When the gas expands, its pressure falls. Makes sense? So the pressure form is: minus two point three zero three, n R T, log of P one over P two. Initial pressure on top. With me?

3:11Energy: where does the work come from?

Now, back to our puzzle. The little triangle is the capital Greek letter delta, meaning change in, so delta U is the change in internal energy. For an ideal gas, U depends only on temperature, which is constant here. So what is delta U? Zero, exactly! And delta H, the enthalpy change, is also zero, for the same reason. Okay? Now the first law: delta U equals q plus w, where q is the heat absorbed by the gas. Delta U is zero, so q equals minus w. Puzzle solved! The gas takes in exactly as much heat from the surroundings as the work it does. Clear? But one condition: an ideal gas. In a real gas, attractions between molecules make U depend on volume too, so delta U is not zero; the slashed equals sign means not equal to. Got it?

4:05Reversible expansion gives the maximum work

Now look at these three graphs: pressure, capital P, going up, and volume, capital V, across. Each time, the gas expands from V one to V two, and the shaded area is the work done by the gas. Graph a: one step against the final pressure, P two. Just a flat strip: the least work. Okay? Graph b: two steps, first against a higher pressure, then against the final one. More shaded area, so more work. Right? So now tell me. Which path gets the most work out of the gas? The reversible one, yes! Infinitely many tiny steps, so the external pressure follows the curve all the way, and the whole area under it is shaded: n R T times the natural log of V two over V one. Reversible expansion gives the maximum work. Clear? And the other way round: reversible compression needs the minimum work. Maximum work out, minimum work in. With me?

5:06Example 1: w and q when 2 mol triples its volume at 400 K

Alright, a question from the course. Two mol, that is two moles, of an ideal gas at four hundred kelvin, written capital K, triples its volume reversibly and isothermally. Find w and q. Given: the natural log of three is one point zero nine nine. Pause the video and try it. Done? Okay, let's check. Triples its volume means V two over V one is three, and its natural log is given: one point zero nine nine. Right? So w equals minus two, times eight point three one four, that's R in joules per kelvin per mole, times four hundred, times one point zero nine nine. The answer is in joules, capital J, okay? That gives minus seven thousand three hundred and nine point seven joules. The wavy equals sign means approximately: about minus seven point three one kilojoules. Negative, because the gas does the work. Clear? Now for q: ideal gas, constant temperature, so q equals minus w: plus seven point three one kilojoules. There's our opening puzzle in numbers! Seven point three one kilojoules of work out, and the same amount of heat in.

6:22Example 2: final volume when the gas does 2.0 kJ of work

Now a harder one: this time we go backwards. One mole of an ideal gas at three hundred kelvin, in ten litres, capital L, does two point zero kilojoules of work in a reversible isothermal expansion. Find the final volume. Pause and try. Watch the sign of w, okay? Done? Okay, let's check. First the work. Two kilojoules is two thousand joules. So we put w equals plus two thousand, right? Oh no, careful! The gas does the work, so w is minus two thousand joules. With plus two thousand, V two would come out less than ten litres. For an expansion! Impossible. So: minus two thousand equals minus one, times eight point three one four, times three hundred, times the natural log of V two over ten. The minus signs cancel. With me? Divide: the natural log of V two over ten equals two thousand, over eight point three one four times three hundred. That's zero point eight zero two. Okay? Now undo the log: V two equals ten times e to the power zero point eight zero two. The small e is the base of natural logs, and this power is about two point two three. So V two is about twenty two point three litres: bigger than ten, as it should be. That's it!

7:49Quick revision: w = -nRT ln(V2/V1), q = -w, maximum work

So, quick revision. One: w rev equals minus n R T, natural log of V two over V one, or minus two point three zero three n R T, log of P one over P two. Initial pressure on top! Two: ideal gas, constant temperature: delta U and delta H are zero, so q equals minus w. Three: reversible expansion gives the maximum work; reversible compression needs the minimum work. Now try the practice problems on this page. All the best!