Overview

Transcript

0:00Introduction

Hello everyone! Okay, two small questions before we start. One: can a gas get hotter, without any heat being supplied to it? Two: can you supply heat to a gas, without its temperature rising at all? Most of us would say no to both, right? Heat in means hotter, and hotter means heat in. Questions like these come up often in JEE, in statement form. First, how to use the problem bank of this unit on thermodynamics; then we answer both. So, let's begin.

0:34What is in this chapter

So, first, what is this chapter? It is the problem bank for the whole unit on thermodynamics: mixed solved problems, in three parts. Part A is at JEE Main level, okay? Part B has three longer problems, at JEE Advanced level. There is a Stirling cycle, a loop on the p V diagram, that is, pressure p against volume V, and a gas compressed suddenly. We will solve the Stirling cycle today, as Example two. Right? And Part C has four concept checks: the ideas often tested in statement form. Our two questions from the start? Both are Part C checks. Clear?

1:13Part A: a 30-minute test

Now, Part A, the JEE Main part. When should you attempt it? On the first day of the unit, just to have a look? Not on day one! Only after Chapters one to six, once the whole unit is done. Why? These problems mix everything: the first law, processes, cycles, adiabatics, specific heats and engines. So finish every chapter first. With me? And attempt it as a real test: thirty minutes, a clock on the table, and the solutions covered. Solve each problem yourself first, okay? Part A gives you five solved problems, and then two more practice problems on the page. Got it?

1:55Part C: one law answers both questions

Now, our tool is the first law, Q equals delta U plus W, okay? U is the internal energy of the gas, and delta, the little triangle, means change in. With me? The sign convention of this course, said once: Q is the heat given to the gas, and W is the work done by the gas. So heat lost makes Q negative, and work done on the gas makes W negative. Clear? Now, question one. No heat goes in, so Q is zero. Can the gas still get hotter? Yes, it can! Adiabatic compression: squeeze the gas with no heat exchanged at all. See it in the first law. Q is zero, and the gas is compressed, so W, the work done by the gas, is less than zero. So delta U equals minus W, which is greater than zero. The internal energy goes up, so T, the temperature, rises. Got it? Now, question two. Heat goes into the gas. Must its temperature rise? No, it need not! Isothermal expansion: the gas expands at constant temperature while heat flows in. Why? For an ideal gas, U depends on the temperature only. T is fixed, so delta U is zero, and Q equals W: all the heat comes out as work. That's it: two puzzles, one law!

3:13Example 1: delta U for gas compressed at constant pressure

Now, Part A, an easy one. Here it is: a gas is compressed at a constant pressure of two times ten to the power five pascal, P a, from three litres to one litre, capital L, while losing one thousand joules, capital J, of heat. Find delta U, the change in internal energy. Pause the video, and try it first. Done? Okay, let's check. Units first. One litre is ten to the power minus three cubic metres, m cubed. So delta V, the change in volume, is one minus three, times ten to the power minus three: minus two times ten to the power minus three cubic metres. Negative, because the gas is compressed. Right? At constant pressure p, the work done by the gas is p times delta V, okay? Two times ten to the power five, times minus two times ten to the power minus three: minus four hundred joules. Negative: work is done on the gas. The gas loses one thousand joules of heat, so Q is minus one thousand joules. Now, from the first law, delta U is Q minus W, okay? Minus one thousand, minus four hundred: minus fourteen hundred joules, right? Oh no, careful! That is the trap. W itself is minus four hundred. So delta U equals Q minus W, that is, minus one thousand, minus, bracket, minus four hundred. Two minus signs make a plus: minus one thousand plus four hundred, that is, minus six hundred joules. Keep negative values in brackets, and this trap cannot catch you.

4:55Example 2: Stirling cycle efficiency versus Carnot

Now Part B, a hard one. One mole of a monatomic gas goes round a cycle. First, an isothermal expansion at six hundred kelvin, capital K, from V naught, V with a small zero, to two V naught. Then an isochoric cooling, at constant volume, to three hundred kelvin. Then an isothermal compression at three hundred kelvin, back to V naught. And an isochoric heating, back to six hundred kelvin. This is a Stirling cycle: find its efficiency, without regeneration. Pause the video, and try it. Done? Okay, let's check. In an isothermal step of an ideal gas, delta U is zero, so Q equals W equals n R T, times l n of V two over V one. Here n is the number of moles, R the gas constant, l n the natural log, and V two over V one is final over initial volume. In an isochoric step, the volume is fixed, so W is zero. Okay? So only the isothermal steps do work, with n equal to one. Net work: R times six hundred minus three hundred, times natural log two: three hundred R natural log two. Clear? Now, the heat taken in. The isothermal expansion takes in six hundred R natural log two. So the efficiency is three hundred over six hundred: fifty percent. Done, right? No, no, wait! That is the trap. The gas also takes in heat in the isochoric heating, from three hundred to six hundred kelvin, and without regeneration that heat comes from outside. At constant volume, Q is n C v delta T, where C v, with the small v, is the molar heat capacity at constant volume. For a monatomic gas C v is three over two R, so this heat is three over two R, times three hundred. Got it? Six hundred R natural log two works out to four hundred and fifteen point nine R, and three over two times three hundred is four hundred and fifty R, okay? So Q in, the total heat taken in, is eight hundred and sixty five point nine R, right? Efficiency, eta, this Greek letter, is W over Q in. Three hundred R natural log two is two hundred and seven point nine R, and the R cancels. Two hundred and seven point nine over eight hundred and sixty five point nine: about twenty four percent. The wavy equals sign means approximately. With me? Now compare with a Carnot engine between the same two temperatures: one minus T two over T one, T one hot and T two cold, both in kelvin. One minus three hundred over six hundred: fifty percent. Twenty four percent is well below it. Good: by Carnot's theorem, no engine between these two temperatures can beat Carnot. And the fifty percent from the trap? It equalled the Carnot value exactly, which is a warning sign for a cycle like this one, without regeneration. Recheck the heat taken in. Remember that check.

7:56Quick revision: Parts A, B, C, first law and efficiency

So, quick revision. Part A is a thirty minute test, after Chapters one to six. Part B has the longer, JEE Advanced problems, and Part C checks your concepts. And remember: Q equals delta U plus W, with W done by the gas, and efficiency is the net work over all the heat taken in. Now set your clock, open Part A, and begin. All the best!