General Processes and the Highest Temperature

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle first. One mole of gas moves along a straight line on the p V graph, and it starts and ends at the same temperature. So its temperature stayed the same all along, right? Actually, no: in the middle, it gets hotter. Where is the hottest point, and how hot? This comes up often in JEE, so let us begin.

0:25Any process p=f(V): work and temperature

We have met processes at constant volume, pressure and temperature. But often the gas follows some other rule: p equals f of V, okay? Small p is the pressure, capital V the volume, and f of V is just some formula in the volume. Now, the gas expands along this rule. How do we find the work done by the gas? Yes, exactly, we integrate! W equals the integral from V one to V two of f of V, d V, okay? The tall S is the integral sign; V one, with the small one below, is where the gas starts, V two where it ends, and d V a tiny change in volume. It is the area under the curve. Clear? Now the temperature. For an ideal gas, p V equals n R T: n moles, capital R the gas constant, capital T the temperature. The double arrow means so: T equals p V over n R, which is V times f of V, over n R, at every point. Got it? Now the big question of today. Where on the path is the gas hottest? Where V times f of V is greatest! The n and R are fixed, so the gas is hottest where p V is greatest. So set d by d V, the derivative with respect to V, of V times f of V, equal to zero; the square brackets just hold the product. With me?

1:54Straight line: hottest at the midpoint

Now the most common case: a straight line on the p V graph. p equals p naught times, bracket, a minus b times V over V naught. The small zero below is read naught: p naught and V naught are fixed values, and a and b are positive constants, so the line slopes downward. Okay? Now, a quick question. Along this line, the pressure is highest at the small volume end. So the gas is hottest there, right? No, no, careful here, that is the trap! Temperature follows p V, not p alone, and at that end the volume is small. Multiply by V: p V equals p naught times, bracket, a V minus b V squared over V naught; the small two above means squared. That minus sign makes it a downward parabola, like a hill. Clear? Now, where is the top of this hill? p V is zero at V equals zero, and again where the line meets the V axis, since there p is zero. The parabola is symmetric, so its top sits halfway between. Right? So here is the topper's trick. For a straight line, the highest temperature is at the midpoint between its two intercepts. No calculus needed at all. But only if the path reaches that midpoint, so always check that. Got it?

3:16Is heat absorbed? Check the sign of dQ

Next: is heat going in or coming out? The first law says Q equals delta U plus W, okay? Our sign convention: Q is the heat given to the gas, W the work done by it. Delta, the little triangle, means change in, so delta U is the change in internal energy. For one tiny step, d Q equals n C v d T, plus p d V, right? C v, with the small v below, is the molar heat capacity at constant volume, so the first term is the tiny change in U, and p d V the tiny work done. With me? If d Q is greater than zero, heat is absorbed; if less than zero, heat is released. Those pointed signs mean greater than and less than. Clear? So tell me. Somewhere, the gas is expanding and also getting hotter. Is it absorbing heat there? Yes, exactly! Both terms are positive, so d Q is positive. But on our straight line, after the hottest point, the gas expands while it cools: the two terms pull opposite ways. So check the sign at each point, not just once. Got it?

4:32Example 1: p = kV from V1 to 2V1, temperature rise and work

Alright, the first example. The gas goes along p equals k V, from V one to two V one; k is positive, since the pressure is. Show that the temperature rises, and find the work in terms of p one V one, where p one is the starting pressure. Pause the video and try it first. Done? Okay, let's check. T is proportional to p V; the curly sign means is proportional to. And p V is k V squared, which grows as V grows. So the temperature rises! Now the work. The volume changes by V one, and the starting pressure is p one. So the work is simply p one times V one, right? Oh no, careful here, that is the trap! Pressure times change in volume works only at constant pressure, and here it keeps rising. So W is the integral from V one to two V one of k V, d V: k over two, times, V two squared minus V one squared. Okay? Now put V two equals two V one. V two squared is four V one squared. So, k over two, times, four minus one, V one squared: three over two, k V one squared. With me? Last step. At the start, p one equals k V one, so k V one squared is p one V one, and W equals three over two, p one V one. More than p one V one, because the pressure rose up to two p one. Makes sense, right?

6:07Example 2: highest temperature on a straight line p-V path

Now a harder one: our puzzle from the start. One mole of gas goes along p equals p naught times, bracket, two minus V over V naught, from V naught over two to three V naught over two. Find the highest temperature. Pause the video and try it first. Done? Okay, let's check. We want the greatest p V, so multiply the line by V, okay? p V equals p naught times, bracket, two V minus V squared over V naught. Now differentiate with respect to V, and set it equal to zero. p naught times, two minus two V over V naught, equals zero. So V equals V naught. Clear? Now check with the topper's trick. The line meets the V axis at two V naught, and the p axis at two p naught. Midpoint: V naught, p naught. We get the same answer! And V naught lies inside our range, so the gas does pass through it. Got it? So at the hottest point, p is p naught and V is V naught. With one mole, T equals p V over R, okay? So T max, with the small word max below, equals p naught V naught over R, and that is our answer. And the puzzle? At V naught over two, the pressure is three over two p naught; at three V naught over two, half of p naught. The dot means times; both products give three quarters of p naught V naught. Right? So both ends are at three p naught V naught over four R, less than T max. Same temperature at the ends, but hotter in between. Mystery solved!

7:52Quick revision: work, midpoint rule, sign of dQ

So, quick revision. Work is the integral of f of V, d V, and the gas is hottest where p V is greatest. On a straight line, that is the midpoint between the intercepts. For heat, check the sign of d Q at each point. Now try the practice problems on this page. All the best!