Irreversible Work and Free Expansion

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Some ideal gas doubles its volume at the same temperature. How much work does it do? Strange, but it depends! Into empty space: zero work. Against a constant pressure: some work. In two steps: even more work! Same start, same end, three different answers. How? By the end of this lesson, you will know. Questions on this kind of work come up often in JEE, so let's begin.

0:35Irreversible: a finite push

So, what makes a process irreversible? The gas pushes out with its own pressure, and the outside pushes in with the external pressure: capital P with the little subscript e x t, P ext. When these two differ by a finite amount, not a tiny one, the process is irreversible. Okay? Now think. The gas pressure is well above P ext, so the piston shoots out. If you raise the outside pressure by a tiny amount, will the piston come back? No, it won't! The gas pressure is still higher by a finite amount, so the gas keeps expanding. Such a process goes in finite time, and an infinitesimal change cannot reverse it. Clear?

1:16Work against a constant external pressure

Now, the work. With a constant external pressure, w equals minus P ext, times the bracket, V two minus V one. Small w is the work, and V with subscripts one and two are the starting and final volumes. In an expansion, V two is bigger, so w is negative: the gas does the work. Right? Now, a classic trap. The gas pressure keeps falling as it expands. So which pressure goes in here: the gas pressure, or the external pressure? The external pressure, correct! The gas pushes back the outside, so the work depends only on P ext and the change of volume, whatever the gas pressure is. Got it? And a unit fact. Bar times litres gives litre bar, written L bar on screen. One litre bar is exactly one hundred joules; the capital J is joules. Okay?

2:09Several steps: the works add

Now look at these three graphs: pressure P going up, volume V going across. In the first, the gas expands from V one to V two in one step, against the final pressure, P two. The shaded rectangle, P ext times the change of volume, is the work. Right? Now do it in steps. In each step, P ext is constant, so each step gives its own rectangle, and the works add. w equals minus the sum of P ext i times delta V i. The big Greek letter sigma means sum of, delta V i is the change of volume in step i, and the little i numbers the steps. With me? Now the second graph: two steps. Compare it with the first. Which shaded area is bigger? The two-step one, yes! In the first step, the gas pushes against a higher pressure, closer to its own, so it does more work. More steps, more work in expansion. The third graph, with infinitely many steps, is the reversible path: maximum work, written under it as n R T times l n, the natural log, of V two over V one. That is our next lesson. Got it? Compression is the other way round. The outside pressure must stay above the gas pressure all the way, so in one step you push with the highest, final pressure the whole time. More steps, less work needed. Clear?

3:34Free expansion, and isothermal changes

Now, an extreme case. The gas expands into a vacuum: completely empty space, nothing outside at all. How much work does the gas do now? Zero, very good! P ext is zero, and the double arrow means which gives: w equals zero. Nothing pushes back. We call this free expansion. Okay? And for an ideal gas, delta T, the change in temperature, is zero. There are no attractions between the molecules to overcome, so nothing cools the gas. And U and H, the internal energy and the enthalpy, depend only on temperature, so delta U and delta H are both zero. Right? Then the first law gives q equals delta U minus w: zero minus zero. So in a free expansion of an ideal gas, all four, w, q, delta U and delta H, are zero. With me? One last rule for the examples. In an isothermal change of an ideal gas, delta U is zero. So from the first law, q equals minus w: the heat taken in equals the work the gas does. Okay?

4:47Example 1: one-step isothermal expansion, 2 bar to 1 bar

Okay, Example one. One mole of an ideal gas at three hundred kelvin, capital K, expands isothermally from two bar to one bar, in one step, against a constant one bar. Find w and q. Pause the video, and try it first. Done? Okay, let's check. V one equals n R T over P one, the starting pressure: n is the moles, T the temperature, and R the gas constant, zero point zero eight three one four litre bar per kelvin per mole. So that is zero point zero eight three one four times three hundred, over two: twelve point four seven litres; capital L is litres. At one bar, half the pressure, double the volume: twenty four point nine four litres. Got it? Now the work. The gas starts at two bar. So w is minus two times twelve point four seven, right? No, no, careful! Two bar is the gas pressure. The work uses the external pressure: one bar. So w equals minus one, times twenty four point nine four minus twelve point four seven. Clear? That is minus twelve point four seven litre bar, which is minus one thousand two hundred and forty seven joules. Negative: the gas does the work. With me? And q? Isothermal, ideal gas, so delta U is zero, and q equals minus w: plus one thousand two hundred and forty seven joules. That's it!

6:22Example 2: the same expansion in two steps, total work

Now Example two, a little harder. The same expansion of one mole at three hundred kelvin, from two bar to one bar; the arrow means to. But now in two steps: first against one point five bar, then against one bar. Find the total work. Pause and try. Hint: where does step one stop? Done? Okay, let's check. Step one stops when the gas pressure falls to one point five bar. Still three hundred kelvin, so V is zero point zero eight three one four times three hundred, over one point five: sixteen point six three litres. Okay? Step one: against one point five bar, from twelve point four seven to sixteen point six three litres. Minus one point five times the difference: minus six point two four litre bar. Right? Step two: against one bar, from sixteen point six three to twenty four point nine four litres. Minus one times eight point three one: minus eight point three one litre bar. Clear? Now add the two steps: minus fourteen point five five litre bar. That is minus one thousand four hundred and fifty five joules. Okay? Now compare with Example one. One step gave one thousand two hundred and forty seven joules. Is two steps more, or less? More, exactly! The greater-than sign says it: the gas does more work in two steps, just like the bigger shaded area in the second graph. And our puzzle? Same start, same end, different work: the work depends on the path. Got it?

8:05Quick revision: w = -Pext Delta V, steps, free expansion

So, quick revision. One: against a constant external pressure, w equals minus P ext times V two minus V one; never the gas pressure. Two: in steps, the works add; more steps, more work in expansion, less in compression. Three: in the free expansion of an ideal gas, w, q, delta U and delta H are all zero. Now try the practice problems on this page. You can do them. All the best!