The Adiabatic Relations

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take a monatomic gas in a cylinder at three hundred kelvin. Now push the piston in quickly, so that no heat has time to flow, and squeeze the gas to one eighth of its volume. No heat goes in, so the temperature cannot change, right? Well, it changes, and by a lot. Where does that energy come from, and how hot does the gas get? That is today's lesson. Questions on adiabatic processes come up often in JEE, so let's begin.

0:37Adiabatic: no heat in or out

So, start from the first law: Q equals delta U plus W, okay? Our sign convention: capital Q is the heat given to the gas, and W is the work done by the gas. Delta, the little triangle, means change in, so delta U is the change in internal energy. Clear? Now, adiabatic means no heat is exchanged. So what is Q here? And then, what is W? Yes, exactly! Q is zero. The double arrow means so: W equals minus delta U, okay? Whatever work the gas does, it pays for from its own internal energy. Right? How do we get no heat flow? Insulated walls, or a process so fast that heat has no time to flow, like our piston. Clear? Now tell me. The gas expands adiabatically and does positive work. Does it get hotter or cooler? Cooler, correct! W is greater than zero, the pointed sign means greater than, so delta U is less than zero. For an ideal gas, U depends only on temperature, so the temperature falls. With me? Compression is the other way round: W is negative, delta U is positive, and the gas heats up. So in our puzzle, the energy comes from the work done on the gas. Got it?

1:59Three forms of the adiabatic relation

Now the key number: gamma, the Greek letter like a tall y. Gamma equals C p over C v: the molar heat capacities at constant pressure and at constant volume, with the small p and v below. Gamma is five over three for a monatomic gas, and seven over five for a diatomic gas. Okay? Now the main relation. In a reversible adiabatic process, p V to the power gamma stays constant: gamma sits up on the V as a power, and const is short for constant. Clear? But often a question gives temperature and volume, not pressure. What do we do then? Bring in p V equals n R T, where n is the number of moles and capital R the gas constant. Put p equals n R T over V, and one power of V cancels. So T times V to the power gamma minus one is constant. Right? And with pressure and temperature, use the third form: p to the power one minus gamma, times T to the power gamma, is constant. Three forms, one process: pick the one that matches the question. Got it? Quick check. Does p V to the power gamma stay constant in every adiabatic process? Not in every one, so be careful here! It needs a reversible, quasi static process, with the gas in equilibrium at every stage, and it needs an ideal gas. So in JEE, a quick compression means quick enough to stop heat flow, yet smooth enough to stay in equilibrium. With me?

3:35Work done, and the steeper curve

Now the work done by the gas in this reversible process. W equals p one V one minus p two V two, over gamma minus one; the small one below means the start, and the small two, the end. Since p V equals n R T, this is also n R times, T one minus T two, over gamma minus one. Okay? But think. No heat comes in, so where does this work come from? From the internal energy, yes! W equals minus delta U, remember? For an ideal gas, delta U is n C v times the temperature change, so W equals n C v times, T one minus T two. In an expansion T one is bigger, so the work is positive. Clear? Last idea: the slope on the p V graph. d p over d V is the slope, and the small letters a d below the bracket mean adiabatic. The adiabatic slope is minus gamma times p over V, right? Now compare. The isothermal curve through the same point, with the small letters i s o below for isothermal, has slope minus p over V, okay? So which curve is steeper? The adiabatic curve, the adiabat, exactly! It is gamma times steeper, and gamma is more than one. Why? In an adiabatic expansion the gas also cools, so its pressure falls faster. Got it?

5:02Example 1: gas compressed to 1/8 volume, final T and p ratio

Alright, examples. This first one is our puzzle. We have a monatomic gas at three hundred kelvin, compressed adiabatically to one eighth of its volume. Find its final temperature, and the factor by which its pressure rises. Pause the video and try it first. Done? Okay, let's check. Monatomic, so gamma is five over three, and gamma minus one is two over three. Okay? Temperature and volume, so we use T V to the power gamma minus one. T two equals T one times, V one over V two, to the power gamma minus one. V one over V two is eight, so T two is three hundred times eight to the power two over three. Right? Eight to the power one third is the cube root of eight, which is two, and two squared is four. So T two is twelve hundred kelvin. Four times hotter, with no heat given at all! Puzzle solved. Now the pressure. The volume became one eighth, so the pressure becomes eight times, right? Oh no, careful, that is the trap! Eight times is the isothermal answer. Here the gas heats up too, so the pressure rises even more. Use p V to the power gamma: p two over p one is eight to the power five over three, that is two to the power five, thirty two. The pressure rises thirty two times!

6:27Example 2: 2 moles expand to twice the volume, work done

Now a harder one. Two moles of monatomic gas at three hundred kelvin expand adiabatically to twice their volume. Find the work done by the gas. Pause the video and try it first. Hint: find the final temperature first. Done? Okay, let's check. Same relation, gamma minus one is two over three, and now V one over V two is one half. T two equals three hundred times, one half to the power two over three, that is three hundred times two to the power minus two over three. The wavy equals sign means approximately: about one hundred and eighty nine kelvin. Expansion cools the gas, see? Now the work: W equals n C v times, T one minus T two, with C v equal to three over two R for a monatomic gas, okay? Now, the temperature falls, so delta U is negative. So the work is negative too, right? No, no, careful here! W equals minus delta U, so a fall in internal energy means positive work, as it must be for an expansion. So W equals two, times three over two, times eight point three one four, the value of R, times, three hundred minus one eighty nine. With me? That gives about two thousand seven hundred and sixty nine joules, or about two point seven seven kilojoules. That is our answer. Got it?

7:52Quick revision: Q = 0, the three relations, steeper adiabat

So, quick revision. Adiabatic means Q is zero, so W equals minus delta U: expansion cools, compression heats. For a reversible adiabatic process of an ideal gas, these three quantities stay constant. The work is n C v times, T one minus T two, and the adiabat is gamma times steeper than the isotherm. Now go and try the practice problems on this page. All the best!