Reversible Adiabatic Processes

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Pump up a bicycle tyre quickly, then touch the pump. It's warm! No flame, no heater. So where did that warmth come from? Hmm. Hold that thought. This idea comes up often in JEE questions. So, let's begin.

0:22Adiabatic: no heat in or out

So, what does adiabatic mean? No heat enters or leaves the gas. Small q is the heat absorbed by the gas, so q equals zero. Okay? Now the first law: delta U equals q plus w. Delta U is the change in internal energy; the little triangle is the capital Greek letter delta. Small w is the work done on the gas. Put q equal to zero. What is left? Delta U equals w, exactly! All the work done on the gas goes into its internal energy, capital U, with nothing lost as heat. Clear? Now you tell me. You compress the gas. Does its temperature, capital T, go up or down? Up, yes! The sign means greater than: w is greater than zero, so U rises, and for an ideal gas, so does the temperature. And there's our bicycle pump! Quick strokes leave no time for heat to escape, so your work warms the air. Got it? Expansion is the opposite: w is less than zero, so U and T both fall. Adiabatic expansion cools the gas. With me?

1:30The adiabatic laws: reversible changes only

First, from the heat capacity lesson. Gamma, the Greek letter on screen, is C P over C V: the molar heat capacities at constant pressure and at constant volume; the P and V are small subscripts, okay? Monatomic: five over three; the wavy sign means about, one point six seven. Diatomic: seven over five, one point four zero. Right? For a reversible adiabatic change of an ideal gas, P V to the power gamma is constant: pressure, times volume raised to the power gamma. Three conditions: reversible, adiabatic, ideal gas. Clear? But many questions give temperature and volume instead. How do we get a law in those two? The ideal gas equation, yes! P equals n R T over V, with n the number of moles and R the gas constant. Put it in, and n R comes out in front. Since n R is fixed, T V to the power gamma minus one is constant. Got it? The same way, in temperature and pressure: T to the power gamma, times P to the power one minus gamma, is constant. Pick the form that matches the data. Okay? Quick test. An ideal gas in an insulated cylinder expands suddenly against a constant outside pressure. So P V to the power gamma is constant, right? No, no, careful! Classic trap. These laws hold only for reversible changes. For a sudden one, use the first law: n C V delta T equals minus P ext delta V, where P with the subscript e x t is the outside pressure. With me?

3:12Adiabat versus isotherm

Look at this graph: pressure up, volume across. Both curves start from the same state: P one, V one, T one, each with a small subscript one. The blue curve is isothermal: P V is constant, and T two equals T one. The orange curve is adiabatic: P V to the gamma is constant, and T three is less than T one. So which curve falls faster? And why? The orange adiabat, correct! At the starting point, where both curves begin, its slope is gamma times the slope of the isotherm. Okay? And why? On the isotherm, the pressure falls only because the volume grows. On the adiabat, the volume grows and the gas also cools, so the pressure falls for two reasons. Right? Now, the work done by the gas is the area under the curve. Up to the same final volume, V two, the isotherm stays higher, so its area is bigger. So the gas does more work isothermally. And as the figure notes, delta U is zero on the isotherm, but negative on the adiabat. Clear?

4:21Work in a reversible adiabatic change

Now, the work. Since q is zero, w equals delta U, which is n C V delta T for an ideal gas. So w equals n C V, times T two minus T one, in brackets. Okay? And from pressures and volumes: w equals P two V two minus P one V one, all over gamma minus one. Both agree; we'll check that soon. For a compression, w comes out positive. Got it?

4:49Example 1: final temperature after compressing to V/8

Alright, a course question. So, a monatomic ideal gas at twenty seven degrees Celsius, the little circle and capital C, is compressed reversibly and adiabatically to one eighth of its volume. Find the final temperature. Pause the video, and try it first. Done? Okay, let's check. The gas starts at twenty seven, so T one is twenty seven, right? Oh no, careful! Gas laws need kelvin. Add two hundred and seventy three: T one is three hundred kelvin, capital K on screen. Okay? Temperature and volume, so T V to the power gamma minus one stays constant. T two equals T one, times V one over V two in brackets, to the power gamma minus one. Monatomic, so gamma minus one is five over three minus one, two over three. And one eighth of the volume means V one over V two is eight. Right? So T two is three hundred, times eight to the power two over three. Topper's trick: eight is two cubed, so its cube root is two, and two squared is four. T two is twelve hundred kelvin, four times the start! Our bicycle pump again, with a much bigger squeeze. That's it!

6:08Example 2: final pressure and work done, by both formulas

Now a harder one. One mole, that's mol on screen, of a monatomic gas at three hundred kelvin and one bar, filling twenty four point nine litres, capital L on screen. It is compressed reversibly and adiabatically to one eighth of its volume. Find P two, and the work done on it, by both formulas. Pause and try. Hint: which law has pressure and volume? So, P two first. One eighth the volume, so eight times the pressure: eight bar, right? No, no, wait! That is Boyle's law, only for constant temperature, and here the gas heats up. So use P V to the power gamma: P two equals P one, times V one over V two, to the power gamma. One times eight to the power five over three. Same trick: eight is two cubed, so this is two to the five, thirty two bar. With me? Now V two: twenty four point nine over eight, three point one one litres. Okay? First formula: thirty two times three point one one, minus one times twenty four point nine, all over two over three. That is one hundred and eleven point nine litre bar. Right? Now in joules, capital J: one litre bar is one hundred joules. So w is about eleven point two kilojoules, the k J on screen. Positive: work is done on the gas. Clear? Now the second formula, n C V delta T, needs T two. Must we work it out again? No! Same gas, same start, same squeeze: that's Example one, so T two is twelve hundred kelvin. For a monatomic gas, C V is three over two R, and R is eight point three one four, okay? So w is one, times three over two, times eight point three one four, times twelve hundred minus three hundred. That's about eleven point two kilojoules. Both formulas agree. Perfect!

8:13Quick revision: q = 0, w = Delta U, the adiabatic laws

So, quick revision. One: adiabatic means q is zero, so w equals delta U, n C V times T two minus T one. Two: P V to the gamma, and its two other forms, stay constant only in reversible changes. Three: compression warms, expansion cools, and the adiabat is steeper. Now go and try the practice problems on this page. All the best!