Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. You push a piston in and squeeze a gas, so you are doing work on it, putting energy in. So its internal energy must go up, right? Not always, and that is the fun part. Today you will see a squeezed gas whose internal energy goes down, by six hundred joules. This is part A of our unit problem bank, at JEE Main level: three quick tools, and four problems from the bank to use them on, each with its own small trap. So, let's begin.

0:33Tool 1: the first law and its signs

So, tool one, the first law: Q equals delta U plus W, okay? Delta U, the triangle you see is the Greek capital delta, means the change in internal energy. The formula is easy; the signs are where students lose marks. So, a quick question on signs. If the gas loses heat, is Q positive or negative? Negative, yes, exactly! In this course, Q is the heat given to the gas. So heat lost means Q is less than zero; that is the less-than sign on screen. Right? Now, if the gas is squeezed, what is the sign of W? Negative again, correct! W is the work done by the gas. Compress it, and the volume falls, so the gas does negative work. So, our sign convention: heat given to the gas, work done by the gas. Okay? And when the pressure stays constant, the work is simply small p, the pressure, times delta V, the change in volume, final minus initial. With me?

1:37Tool 2: Cp, Cv and gamma from Mayer's relation

Tool two, with a quick one from the bank. For a certain gas, C v is twenty joules per mole per kelvin. Find gamma, the Greek letter at the end. C v, capital C with a small v below, is the molar heat capacity at constant volume. Okay? First, Mayer's relation. For an ideal gas, per mole, C p minus C v equals R, the gas constant. C p, with a small p below, is the molar heat capacity at constant pressure. Clear? Now think. Why should C p be bigger than C v? Yes, exactly! At constant pressure the gas expands and does work, so part of the heat goes into work, not into the temperature. So, for each kelvin of rise, one mole needs R more heat. And gamma is the ratio, C p over C v. Okay? So, C p is twenty plus eight point three one four: twenty eight point three one four, about twenty eight point three joules per mole per kelvin; the wavy equals sign means approximately. Right? And gamma is twenty eight point three one four over twenty: about one point four two. The trap here is writing C v over C p, and getting less than one. Since C p is C v plus R, gamma is always bigger than one. Got it?

2:56Tool 3: the heat engine

Tool three: the heat engine. An engine of efficiency thirty percent rejects seven hundred joules per cycle. Find the heat absorbed and the work done. Look at the figure. The engine takes heat Q one from the hot reservoir at temperature T one, and rejects Q two to the cold one at T two; the small one and two below are labels. The difference comes out as work: W equals Q one minus Q two. On the right are two formulas: the efficiency, which we build in a moment, and the Carnot one, with temperatures in kelvin, which is only for an ideal engine; we do not need it today. Okay? Efficiency is eta, the Greek letter that looks like an n with a long tail. It is the work out over the heat in: W over Q one, or one minus Q two over Q one. Right? Now, thirty percent efficient, seven hundred joules rejected. So the heat absorbed is seven hundred over zero point three, right? The slash on screen means over. Oh no, careful here, that is exactly the trap! Thirty percent of Q one becomes work; the seven hundred joules thrown away is the other seventy percent. So Q two is seventy percent of Q one. Clear? So Q one is seven hundred over zero point seven: one thousand joules of heat absorbed. With me? And the work is Q one minus Q two: one thousand minus seven hundred, three hundred joules. Quick check: three hundred out of one thousand is thirty percent. Perfect!

4:25Example 1: delta U of a gas compressed while losing heat

Now, example one: our puzzle from the start. Here, a gas is compressed at constant pressure, two times ten to the power five pascal, from three litres to one litre, while losing one thousand joules of heat. Find delta U, the change in internal energy. Capital P small a is pascal, and capital L is litre. Pause the video and try it first. Done? Okay, let's check. Change in volume: one minus three, minus two litres. One litre is ten to the power minus three cubic metres, so delta V is minus two times ten to the power minus three cubic metres. Right? So W equals p delta V: two times ten to the power five, times minus two times ten to the power minus three, in brackets. Minus four hundred joules. Negative, because the gas is compressed. Okay? Now the heat. The gas loses one thousand joules, so Q is plus one thousand, right? No, no, careful here, remember our signs! Q is the heat given to the gas. This gas loses heat, so Q is minus one thousand joules. Got it? Now the first law gives delta U equals Q minus W, okay? Minus one thousand, minus minus four hundred: that is minus one thousand plus four hundred, minus six hundred joules. And that solves our puzzle. We did four hundred joules of work on the gas, but it lost one thousand joules as heat, so its internal energy fell by six hundred joules. See?

5:58Example 2: temperature rise in adiabatic compression of air

Now example two, a little harder. Air, with gamma equal to one point four, as the brackets say, is compressed adiabatically, so that its pressure doubles. By what factor does its temperature rise? Adiabatic means no heat flows in or out. Pause the video and try it first. Done? Okay, let's check. Here is the first instinct: the pressure doubles, so the temperature doubles too, right? Oh no, careful here, that is true only at constant volume! Here the volume shrinks as well. So we use the adiabatic relation: p to the power one minus gamma, times T to the power gamma, is constant. It holds only for a reversible adiabatic process of an ideal gas, and we treat the air that way. Okay? Rearrange it. T two over T one equals p two over p one, in brackets, to the power gamma minus one over gamma. T one and p one, with a small one below, are the start; T two and p two are the end. Clear? Now, gamma is one point four. What is the power? Yes, very good! Zero point four over one point four, which is two over seven, okay? So T two over T one is two to the power two over seven, about one point two two. So the temperature rises by a factor of about one point two two, not two: adiabatic compression heats the gas. Got it?

7:27Quick revision: first law signs, gamma, engine, adiabatic

So, quick revision. First law: Q equals delta U plus W, and heat lost or a compression gets a negative sign. C p minus C v equals R per mole, and gamma, C p over C v, is greater than one. For an engine, eta is W over Q one, and Q one equals W plus Q two. And for a reversible adiabatic process, T two over T one is p two over p one, to the power gamma minus one over gamma. Now go and try the practice problems on this page. All the best!