Mixed Problems: Set 3

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. Take an ideal gas, and let it double its volume, ending at its starting temperature. Let it expand slowly, keeping its temperature fixed, and the gas does about zero point six nine n R T of work. Let it expand suddenly, against a fixed outside pressure equal to its final pressure, and it does only half n R T of work. Same gas, same start, same end. So how can the work be different? We will settle it in the very first idea. This is set three of mixed problems on thermodynamics, the kind that come up often in JEE Main. So, let's begin.

0:41Idea: same start, same end, different work

So, our puzzle first. Here, an ideal gas expands isothermally, at a fixed temperature T, from volume V to two V, against a constant external pressure equal to the final pressure. Find the work done, and compare it with the reversible isothermal work. Okay? First, what is that final pressure? The gas ends at twice the volume, at the same temperature. So what does p V equals n R T give you? Yes, exactly! n R T over two V, with n the number of moles and R the gas constant. That is the outside pressure here, p ext: small p, with e x t written below, for external. Right? The outside pressure stays constant, so the work done by the gas, W, is p ext times the change in volume. Two V minus V is just V, okay? So n R T over two V, times V, gives half n R T of work. Clear? Now let the same gas expand the slow way, reversibly, at the same temperature. Will it do more work, or less? It does more work, yes! For a reversible isothermal expansion, W equals n R T times the natural log of V two over V one, written l n. V one and V two, with a small one and a small two below, are the starting and final volumes. V two over V one is two, and the natural log of two is about zero point six nine; the wavy equals sign means approximately. With me? So half n R T is less than zero point six nine n R T; the pointed sign means less than. The sudden, irreversible expansion gives less work. Got it? And that settles our puzzle. Work W, and heat Q too, depend on the path, not just on where the gas starts and ends. But the change in internal energy, delta U, does not; the triangle is the Greek capital delta, for change. Here the temperature ends where it started, so for an ideal gas, delta U is zero both ways. Okay?

2:51Example 1: heat absorbed along a straight line p-V path

Alright, example one. One mole of monatomic gas goes from state A, at p naught and V naught, to state B, at two p naught and three V naught, along a straight line. p naught and V naught, with a small zero below, are a starting pressure and volume, and each pair of brackets gives the pressure first, then the volume. Find the heat absorbed. Pause the video, and try it first. Done? Okay, let's check. First, the work done by the gas: it is the area under its p V graph, and under a straight line that area is a trapezium. Half the sum of the two pressures, p naught plus two p naught, times the width, three V naught minus V naught. Half of three p naught, times two V naught: three p naught V naught. Right? Now delta U, and for that we need the change in temperature. The pressure doubles, so the temperature doubles too, right? Oh no, careful here, that is the trap! The volume also triples. From p V equals n R T, the temperature follows the product of p and V, okay? So p AVA, with a small A below for state A, is p naught V naught; and p BVB is two times three, six p naught V naught. The temperature goes up six times, not two. With me? Monatomic, so C v, capital C with a small v below, the molar heat capacity at constant volume, is three over two R, right? Now delta U is n C v delta T, and n R delta T is just the change in p V, written delta, then p V in brackets. So delta U is three over two, times six minus one, p naught V naught: seven point five p naught V naught. Clear? Now the first law, Q equals delta U plus W, with this course's sign convention: Q is the heat given to the gas, and W is the work done by the gas. Seven point five plus three: ten point five p naught V naught of heat absorbed. See, we never needed a single temperature! Got it?

5:04Example 2: efficiency of a rectangular cycle, diatomic gas

Now a harder one: a diatomic gas goes round a cycle. Isobaric expansion, at constant pressure, from p naught, V naught, to two V naught. Then isochoric cooling, at constant volume, down to p naught over two. Then isobaric compression back to V naught, and isochoric heating back to the start. Find the efficiency. Pause the video and try it. Hint: sketch the cycle. Done? Okay, let's check. Two constant pressure steps and two constant volume steps make a rectangle on the p V graph. The net work is its area: the height, p naught minus p naught over two, times the width, two V naught minus V naught. That is half p naught V naught. Right? For the efficiency, we need the heat absorbed. So, which steps take heat in? Only the expansion? Not only the expansion; careful here! In these four steps, heat goes in wherever the temperature rises, and the temperature follows p V, remember? In the isobaric expansion, p V goes up, from p naught V naught to two p naught V naught. And in the isochoric heating, it goes up again, from half p naught V naught back to p naught V naught. In the other two steps, the gas cools and gives heat out. Okay? The expansion is at constant pressure, so we use C p, with a small p below: seven over two R for a diatomic gas, right? Since n R delta T is the change in p V, Q is seven over two times that change, one p naught V naught. So this step absorbs seven over two p naught V naught. With me? The heating is at constant volume, so we use C v, five over two times R, and p V rises by half p naught V naught. Five over two, times one half: five over four p naught V naught. Clear? So the total heat absorbed, Q in, with in written small below, is seven over two plus five over four: nineteen over four p naught V naught. Okay? The efficiency, eta, the Greek letter like an n with a long tail, is the work over the heat absorbed. Half, over nineteen over four: two over nineteen, about ten point five percent. And if you forget the isochoric heating, your efficiency comes out too high. Got it?

7:30Quick revision: path dependence, work, heat and efficiency

So, quick revision. Work and heat depend on the path; delta U does not. Against a constant external pressure, W is p ext times the change in volume; that is less work than the reversible isothermal expansion between the same states. For an ideal gas, n R delta T is just the change in p V, so delta U equals n C v delta T needs no temperatures. And for a cycle, efficiency is the work over the heat absorbed; count every step that takes heat in. Now go and try the practice problems on this page. All the best!