Part C: Concept Checks

Transcript

0:00Introduction

Hello everyone! Okay, two small puzzles before we start. One: can a gas get hotter, without you giving it any heat at all? Two: can you keep giving a gas heat, while its temperature does not rise even one degree? Sounds impossible, right? Heat in should mean hotter. Hold on to your answers. Today: four quick concept checks on thermodynamics. Such one line reasoning questions come up often in JEE, so let's begin.

0:31The tools: the first law and internal energy

So, first our tool, the first law: Q equals delta U plus W, okay? Our sign convention: capital Q is the heat given to the gas, and capital W is the work done by the gas. U is the internal energy, and delta, the little triangle, means change in. Internal energy is a state function: it depends only on the state of the gas. But Q and W are path functions: they depend on the process. Right? Now tell me. For an ideal gas, what does the internal energy depend on? On temperature alone, yes! U equals f over two, n R T, okay? Here f is the number of degrees of freedom, n the number of moles, R the gas constant, and capital T the absolute temperature. Clear? So for an ideal gas, delta U greater than zero means the temperature rises, and delta U less than zero means it falls; the pointed signs mean greater than and less than. With me?

1:29Check 1: is U of a real gas a function of T alone?

Now, check one. Is the internal energy of a real gas also a function of temperature alone? Many students say yes straight away. Are they right? Look at what internal energy really is. U is the total kinetic energy plus potential energy of the molecules. The kinetic part depends on the temperature, and that is all it depends on. Okay? But in a real gas, the molecules pull on each other: these are intermolecular forces. The double arrow means so: the potential energy depends on how far apart the molecules are. Right? And how far apart they are depends on the volume, capital V, okay? So the internal energy of a real gas depends on the volume, as well as on the temperature. Clear? So the answer is no! In an ideal gas, we assume no forces between the molecules, except during collisions. Only for an ideal gas is U equal to U of T: a function of temperature alone. That is why we keep saying ideal gas today. Got it?

2:35Check 2: hotter, with no heat supplied?

Check two, our first puzzle. In the figure, small p, the pressure, goes up, and capital V, the volume, goes across. Four processes start from the same state, p naught, V naught; the small zero below means starting. The flat line is isobaric, constant pressure; the vertical line is isochoric, constant volume. Okay? The dark blue curve is isothermal, p V equals constant. The orange curve is adiabatic, p V to the power gamma equals constant; gamma, the Greek letter, is C p over C v. This holds only for a reversible adiabatic process of an ideal gas. Right? Now, the puzzle. Can the temperature of a gas rise, without any heat being supplied? Most students say no: no heat, no change. Is that correct? Oh no, careful here! Use the first law. Adiabatic means no heat is exchanged, so Q is zero. So W equals minus delta U: any work done by the gas comes out of its own internal energy. Okay? Now compress the gas: we push the piston in and do work on it. On the figure, that is the orange curve run backwards, from right to left. The work done by the gas is negative, W less than zero, so delta U is greater than zero. With me? And for an ideal gas, U depends on temperature alone. So delta U greater than zero means the temperature rises. Clear? So, yes! Adiabatic compression heats the gas, with no heat supplied at all. The energy came in as work, not as heat. Puzzle one, solved!

4:16Example 1: heat in with no temperature rise, isothermal

Now, remember our second puzzle? Here it is, as Example one. Can heat be supplied to a gas without its temperature rising? Hint: pick the right curve on the figure. Pause the video, and try it first. Done? Okay, let's check. Take the dark blue curve, the isothermal one: the temperature stays constant. For an ideal gas, U depends on temperature alone, so delta U is zero. Right? Now the first law: Q equals delta U plus W, that is, zero plus W, so Q equals W, exactly. Okay? For a slow, reversible isothermal process, Q equals W equals n R T times the natural log of V two over V one; l n on screen is the natural log. V one and V two, with a small one and a small two below, are the starting and final volumes. In an expansion, V two is bigger than V one, so this is greater than zero: heat really goes in. Clear? So, yes! In an isothermal expansion, heat goes in, but the temperature does not rise. All the heat becomes work. Puzzle two, solved too! But wait. All the heat becomes work. Does that break the second law? No, it does not! The Kelvin Planck statement says no cyclic process can convert heat completely into work: an engine working in a cycle must always reject some heat. Here the gas does not come back to its start; it ends at a bigger volume. It is not a cycle, so there is no clash with Kelvin Planck. Got it?

5:56Example 2: why a Carnot engine cannot be built in practice

Now Example two, the last check, and a little harder. Why is a Carnot engine impossible to build in practice? Hint: look at what each of its four steps demands. Pause the video, and try it first. Done? Okay, let's check. Many students say: because its efficiency would be one hundred percent, and the second law forbids that. Is that right? No, no, careful here! Its efficiency, eta, the Greek letter like an n with a long tail, is one minus T two over T one; the word Carnot below eta just names it. T one is the hot source and T two the cold sink, both in kelvin. T two is above absolute zero, so eta is less than one, never one hundred percent. So the problem is not the law. Okay? Now the cycle in the figure: small p up, capital V across. From A to B, an isothermal expansion at T one, absorbing Q one. From B to C, adiabatic. From C to D, isothermal at T two, rejecting Q two. And from D back to A, adiabatic again. With me? Take the isothermal steps first, the orange and green curves. They must be infinitely slow, and completely frictionless. Such an engine would take forever to finish even one cycle! Right? Now the adiabatic steps, the two dark blue curves. No heat may leak in or out, so the walls must be perfectly insulated. Clear? But every real process has some friction, runs at a finite speed, and leaks some heat. Real processes are irreversible, while all four Carnot steps must be reversible. Got it? So a Carnot engine is an ideal limit, not a real machine. By Carnot's theorem, it sets the best efficiency any engine can have between those two temperatures, and real engines always fall short of it. That's it, all four checks done!

7:57Quick revision: real gas U, adiabatic, isothermal, Carnot

So, quick revision. Only an ideal gas has U as a function of temperature alone; for a real gas, U depends on the volume too. Adiabatic compression raises the temperature with Q equal to zero; isothermal expansion gives Q equal to W, with the temperature fixed. And a Carnot engine needs reversible steps, but real processes are irreversible. Now go and try the practice problems on this page. All the best!