Part B: Harder Problems

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. One mole of an ideal gas sits in a perfectly insulated cylinder. It expands suddenly, against a constant outside pressure. Because of the insulation, no heat goes in or out. So q, the heat, is zero. And entropy change is heat over temperature. So the entropy change is zero, right? Hmm, not so fast. By the end of this video, you will calculate it yourself, and the answer may surprise you. Today is Part B of our problem bank: harder problems that mix two or three ideas, and in JEE, problems like these come up often. Three tools first, then two problems. So, let's begin.

0:46Tool 1: state functions ignore the path

So, tool one, the first law. For a closed system, delta U equals q plus w. The little triangle is the Greek letter delta, meaning change in; U is the internal energy, q the heat, and w the work. Okay? Now the signs. q greater than zero, the sign on screen, means heat absorbed by the system. And w greater than zero means work done on the system. So heat given out, and work done by the system, are negative. Right? Now the key idea for today. U, H, S and G, that is internal energy, enthalpy, entropy and Gibbs energy, are state functions. So are P, V and T: pressure, volume and temperature. Their change depends only on the start and the end, not on how you got there. With me? Now tell me. Two different routes, between the same two states. Is the heat the same along both? No, not necessarily! q and w are path functions: they depend on the route taken. Only their sum, q plus w, which is delta U, stays the same. Clear? And one special route: adiabatic, with no heat in or out. Then q is zero, so delta U equals w. Got it?

2:07Tool 2: entropy needs a reversible path

Now tool two: entropy, the letter S, also a state function. Delta S equals q rev over T, okay? q rev, with the small r e v below it, is the heat taken in reversibly, at a constant temperature T, in kelvin. If T changes along the path, we need a formula like the one in tool three. Not just any heat: reversible heat. Remember that. So for a reversible adiabatic change, q rev is zero, and delta S is zero. The entropy stays constant. Right? But what if the change is not reversible, like the sudden expansion in our puzzle? Then the real heat is not q rev. So how do we find delta S? With the state function idea, exactly! Delta S depends only on the start and the end. So take any reversible path between the same two states, and calculate along it. With me?

3:02Tool 3: delta S of an ideal gas in two steps

So here is an easy reversible path for an ideal gas, in two steps. Step one: heat the gas at constant pressure, from T one to T two, the initial and final temperatures. Delta S equals n C P, times l n of T two over T one. Here n is the number of moles; C P, the C with a small P below it, is the molar heat capacity at constant pressure; and l n is the natural log. Okay? Step two, isothermal: keep the temperature fixed, and change the pressure from P one to P two. If the pressure falls, does the entropy rise or fall? It rises, correct! The gas spreads into a bigger volume. Delta S equals n R, times l n of P one over P two, where R is the gas constant. Flip the fraction inside the log, and the sign flips: minus n R, times l n of P two over P one. Right? Now add the two steps. Delta S equals n C P l n T two over T one, minus n R l n P two over P one. S is a state function, so this works whether the real change is reversible or not. Clear? And for a monatomic ideal gas, C P is five over two times R: twenty point seven nine joules per kelvin per mole. The powers of minus one on screen mean per kelvin and per mole. Got it?

4:30Example 1: work on an adiabatic path between the same states

Now, example one. Here, a system goes from state A to state B by path one, with q equal to one hundred joules, and w equal to minus forty joules. What is w along an adiabatic path two, between the same two states? Pause the video and try it. Done? Okay, let's check. Path one: delta U equals q plus w, that is one hundred plus minus forty: sixty joules, the capital J on screen. Clear? Now path two. Work along path one was minus forty joules. Same start, same end, so w along path two is also minus forty joules, right? No, no, careful! That is the trap. Work is a path function: it changes with the route. Path two is adiabatic, so q is zero. Okay? And internal energy is a state function. So delta U is sixty joules again, exactly as on path one. With me? So sixty equals zero plus w, and the double arrow gives w equals plus sixty joules. Positive: work is done on the system. On path one the system did forty joules of work; on path two, sixty joules is done on it. Same states, different work. That's it!

5:50Example 2: delta S of an irreversible adiabatic expansion

Now our puzzle from the start, with real numbers. One mole of a monatomic ideal gas, at three hundred kelvin and ten bar, expands adiabatically against a constant one bar, until its pressure is one bar. It ends at one hundred and ninety two kelvin. Find delta S of the gas, and comment. Pause and try. Hint: entropy is a state function. Done? Okay. Adiabatic, so q is zero. And delta S is heat over temperature, so delta S is zero, right? No, no, careful! This is the trap from our puzzle. The expansion is sudden, against a fixed outside pressure, so it is not reversible. q is zero, but this q is not q rev. Okay? So use tool three, a reversible path between the same two states. One mole, so n equals one: delta S equals C P l n T two over T one, minus R l n P two over P one. Right? Now the numbers. C P is twenty point seven nine, R is eight point three one four, and T two over T one is one ninety two over three hundred: zero point six four. P two over P one is one over ten, and minus R l n of one over ten is plus R l n ten. Clear? Now the two terms. The gas cooled, which lowers its entropy, but it also expanded, which raises it. Which effect wins? The expansion, yes! Twenty point seven nine times l n zero point six four is minus nine point two eight. Eight point three one four times l n ten is plus nineteen point one four. With me? So delta S is plus nine point nine joules per kelvin. Puzzle solved! Positive, although q is zero, because the process is irreversible. Zero heat does not mean zero entropy change. Got it?

7:52Quick revision: state functions and entropy change

So, quick revision. One: U and S are state functions, so only the start and the end matter. Two: q and w depend on the path, and adiabatic means q is zero, so delta U equals w. Three: delta S is q rev over T, so for an irreversible change, use a reversible path. Four: for an ideal gas, delta S equals n C P l n T two over T one, minus n R l n P two over P one. Now go and try the other fourteen problems on this page, each one yourself before you read its solution. All the best!