Overview

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take a Carnot engine working between two hundred and twenty seven degrees Celsius and minus twenty three degrees Celsius. One student puts these numbers straight into the efficiency formula, and gets about one hundred and ten percent. An engine that gives out more work than the heat it takes in? Something is wrong here, right? We will find the mistake today. And this video is your map for the mixed thermodynamics problems: what JEE Main asks, and one tool for each type of question. So, let's begin.

0:37How this chapter is built

So, first: JEE Main questions on thermodynamics are short. Okay? They test the first law along a given p V path: small p is the pressure, capital V the volume, and the dash means a graph of p against V, and then the standard processes. Right? Then the adiabatic relations. C p and C v, capital C with a small p or v below, are the molar heat capacities at constant pressure and at constant volume, and gamma, the Greek letter, is their ratio. And Carnot efficiencies. Clear? Now, where are the real exam questions? The JEE Main questions from twenty twenty one to twenty twenty six are at the end of chapters one to five; each dash on this line means to. With me? This chapter is different: its problems are original, written in the same style. Set one is easy, set two medium, and set three hard. Got it? And one value to take: capital R, the gas constant, is eight point three one four joules per mole per kelvin. Capital J is the joule, mol is the mole, capital K is the kelvin, and each small minus one means per.

1:54Tool 1: the first law

Okay? Now, your toolkit. Tool one, the first law: capital Q equals delta U plus capital W, where U is the internal energy, and delta, the little triangle, means change in. Right? Signs matter, so here is the course's convention. Q is the heat given to the gas, and W is the work done by the gas. Clear? And W is the area under the p V curve: positive when the gas expands, and negative when it is compressed. With me? Now think. Two different paths join the same two states. Is delta U the same on both? Yes, exactly! U depends only on the state, and for an ideal gas, on the temperature alone. So delta U equals n C v delta T on every path: n is the number of moles, and delta T the change in temperature. Only Q and W change with the path. Got it?

2:51Tool 2: the standard processes

Tool two. In the graph, the axes are p and V, and all four processes start from the black dot, p nought, V nought, in brackets; the small zeros below mean starting values. The red vertical line is isochoric, constant volume. The volume does not change, so W is zero, and Q equals n C v delta T, okay? The green flat line is isobaric, constant pressure. W equals p delta V, which is n R delta T, right? The dark blue curve is isothermal, constant temperature, where p V is constant, so for an ideal gas, delta U is zero. Then Q equals W equals n R T times the natural log, l n on screen, of V two over V one: the final volume over the starting volume. Clear? And the orange curve, the steeper one, is adiabatic: no heat flows, so Q is zero, and W equals minus delta U, with me? Along this orange curve, p V to the power gamma is constant. But tell me, is that true in every adiabatic process? No, not in every one: only in a reversible adiabatic process of an ideal gas. Got it? The classic trap: free expansion into a vacuum, inside insulated walls. Q is zero, the gas pushes nothing, so W is zero, and so delta U is zero; an ideal gas keeps its temperature. Adiabatic, yes, but not reversible, so p V to the gamma is not constant. Okay?

4:24Tool 3: Cp, Cv, gamma and the Carnot engine

Tool three. Mayer's relation: C p minus C v equals R, per mole of an ideal gas. Right? And gamma is C p over C v: five thirds for a monatomic gas, and seven fifths for a diatomic gas. Clear? Last tool, the Carnot engine. T one, capital T with a small one below, is the temperature of the hot source, and T two that of the cold sink. The efficiency, eta, the Greek letter like an n with a long tail, is one minus T two over T one. With me? Now, can we put T one and T two into this formula in degrees Celsius? No, no, careful here! They must be in kelvin, so add two hundred and seventy three first. Hold on to that; it is the key to our puzzle. Got it?

5:14Example 1: Carnot engine, 227 °C and -23 °C, the 110% trap

Now, back to our puzzle. Example one, an easy one from set one. Find the efficiency of a Carnot engine between two hundred and twenty seven degrees Celsius and minus twenty three degrees Celsius; the small circle before the C means degrees. Pause the video and try it first. Done? Okay, let's check. In the figure, the engine takes heat Q one from the hot reservoir at T one, gives heat Q two to the cold one at T two, and the difference, W, comes out as work. On the right, eta is W over Q one, and for Carnot, one minus T two over T one. First, kelvin: T one is two hundred and twenty seven plus two hundred and seventy three, five hundred kelvin. Right? T two is minus twenty three plus two hundred and seventy three: two hundred and fifty kelvin. Clear? So eta equals one minus two hundred and fifty over five hundred: one half, that is fifty percent. And our puzzle: the student used the Celsius values, one minus, minus twenty three over two hundred and twenty seven, and got about one hundred and ten percent. No engine can do that. So always use kelvin!

6:31Example 2: 1 mole of nitrogen at constant pressure, Q and W

Now a harder one, a medium problem from set two. One mole of nitrogen is heated at constant pressure through twenty kelvin. Find the heat supplied, and the work done. Pause the video and try it first. Done? Okay, let's check. Nitrogen is diatomic, so C v is five halves R, and many students write Q equals n C v delta T: five halves, times eight point three one four, times twenty, about four hundred and sixteen joules; the wavy equals sign means approximately. Is that right? No, no, careful here! C v is for constant volume, but here the pressure is constant. So we need C p. Nitrogen is diatomic, and C p is C v plus R: five halves R plus R, which is seven halves R, okay? So Q equals n C p delta T: one, times seven halves, times eight point three one four, times twenty. About five hundred and eighty two joules. Right? Now the work. At constant pressure, W equals p delta V, which is n R delta T: eight point three one four times twenty, about one hundred and sixty six joules. With me? And that four hundred and sixteen joules from C v is delta U, the rise in internal energy. Check the first law: four hundred and sixteen plus one hundred and sixty six is five hundred and eighty two, the same as Q, so the first law holds. Got it?

8:09Quick revision: first law signs, adiabatic rule, Cp, kelvin

So, quick revision. Q equals delta U plus W, with Q given to the gas and W done by it, and for an ideal gas, delta U depends on temperature alone. p V to the power gamma is constant only for a reversible adiabatic process of an ideal gas. At constant pressure, use C p; and in the Carnot formula, temperatures in kelvin. Now go and try the three sets of this chapter. With this map, you are ready for them. All the best!