Mixed Problems: Set 1

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. Put these two numbers straight into the Carnot efficiency formula, and you get about one point one. That is an efficiency of one hundred and ten percent! An engine that gives out more work than the heat it takes in? Something is wrong here, right? Hold that thought; we will settle it in example one. This is set one of mixed problems on thermodynamics, the short numericals that come up often in JEE Main. So, let's begin.

0:40Idea 1: the first law and its signs

Idea number one: the first law. Here, a gas absorbs two hundred joules of heat, capital J for joules, and its internal energy rises by one hundred and fifty joules. Find the work done by it. Let's think. The first law says Q equals delta U plus W, okay? The triangle is the Greek capital delta, for change, so delta U is the change in the internal energy of the gas. And 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. Now, two hundred joules went in, but only one hundred and fifty stayed inside. So where did the rest go? Into work, yes! W equals Q minus delta U: two hundred minus one hundred and fifty, that is fifty joules. Positive, so the gas did work on its surroundings. Clear?

1:36Idea 2: heat at constant pressure

Idea number two: heat at constant pressure. Here, eight grams of helium, small g for grams, is heated at constant pressure through one hundred kelvin, capital K for kelvin. Find the heat supplied. How many moles is eight grams of helium? Helium's molar mass is four grams per mole; g mol to the power minus one means grams per mole. So n, the number of moles, is eight over four: two moles, not eight! Right? Now, which molar heat capacity do we use? C v, capital C with a small v below, for constant volume, or C p, with a small p, for constant pressure? C p, of course, since the pressure is fixed! So Q equals n C p delta T, where delta T is the rise in temperature, one hundred kelvin. The gas also expands and does work, so it needs more heat than at constant volume. Clear? Helium is monatomic, so C v is three over two R, with R the gas constant. And Mayer's relation, C p minus C v equals R, per mole of an ideal gas, gives C p equals five over two R, okay? Now put it all in, with R as eight point three one four joules per mole per kelvin. Two times five over two, times eight point three one four, times one hundred: four thousand one hundred and fifty seven joules. Got it?

3:07Idea 3: squeeze it fast

Idea number three: adiabatic compression. Here, a monatomic gas at three hundred kelvin is compressed adiabatically, with no heat flowing in or out, to one quarter of its volume. Find its final temperature. Okay? For a reversible adiabatic process of an ideal gas, T V to the power gamma minus one stays constant; const is short for constant. Gamma, the Greek letter, is C p over C v. Right? Monatomic, so gamma is five over three, and gamma minus one is two thirds. Clear? Now, the gas starts at three hundred kelvin, and its volume goes down to a quarter. So do we multiply three hundred by one quarter to the power two thirds, or by four to the power two thirds? By four to the power two thirds! T two equals T one, times the bracket V one over V two, to the power gamma minus one; the small one means before, and the small two means after. V one over V two is four, and compression heats the gas, so the temperature must go up. With me? Four to the power two thirds is the cube root of sixteen, about two point five two; the wavy equals sign means approximately. So T two is about seven hundred and fifty six kelvin. No heat went in, yet the gas got much hotter: the work done on it became internal energy. Got it?

4:34Example 1: Carnot efficiency, 227 C to -23 C in kelvin

Alright, example one: our puzzle from the start. 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 is the degree sign. Pause the video, and try it first. Done? Okay, let's check. The Carnot efficiency, eta, the Greek letter like an n with a long tail, is one minus T two over T one. T one is the hot source, and T two the cold sink. Right? Put the Celsius numbers straight in: one minus, minus twenty three over two hundred and twenty seven, about one point one. One hundred and ten percent? That is our puzzle. So what went wrong? Yes, exactly, the temperature scale! This formula needs absolute temperatures, in kelvin, never in degrees Celsius. Add two hundred and seventy three: T one is five hundred kelvin, and T two is two hundred and fifty kelvin. Okay? So eta is one minus two hundred and fifty over five hundred: one half, or fifty percent. Puzzle solved! The engine was never above one hundred percent; the Celsius numbers fooled us. Got it?

5:54Example 2: work as area under a straight-line p-V graph

Now a harder one. Here, a gas expands along a straight line, from two litres at three atmospheres, to six litres at one atmosphere. Capital L is the litre, a t m is the atmosphere, and each pair of round brackets gives the volume, then the pressure. Find the work done. Pause the video and try it. Hint: draw the graph. Done? Okay, let's check. The work done by a gas is the area under its p V graph: small p, the pressure, against capital V, the volume; the dash means against. The gas expands, so this work is positive. Right? Under a slanted straight line, down to the volume axis, what shape do we get? Yes, a trapezium! Its area is half the sum of the two parallel sides, times the width between them. So W equals half, p one plus p two, times V two minus V one, where one is the start and two is the end. With me? Now the numbers: half of three plus one, that is two atmospheres, times six minus two, that is four litres. So W is eight litre atmospheres. Clear? So the work done is eight joules, right? Oh no, careful here, that is the trap! The litre atmosphere is a unit of energy, but it is not the joule. So convert: one atmosphere is one point zero one three times ten to the power five pascal, written capital P small a, and one litre is ten to the power minus three cubic metre, written m cubed. Okay? So W is eight, times one point zero one three times ten to the power five, times ten to the power minus three. That is about eight hundred and ten joules, not eight! Got it?

7:47Quick revision: first law, adiabatic, Carnot, p-V work

So, quick revision. First law: Q equals delta U plus W, with W done by the gas; and at constant pressure, Q equals n C p delta T, okay? Reversible adiabatic: T V to the power gamma minus one is constant, and compression heats the gas. Carnot: eta equals one minus T two over T one, always in kelvin. And work is the area under the p V graph; turn litre atmospheres into joules. Now go and try the practice problems on this page. All the best!