The Standard Processes

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. You keep giving heat to a gas, yet its temperature does not rise at all. Is that even possible? Hold that thought; by the end, you will know where the heat goes. Questions on these standard processes come up often in JEE, so let's begin.

0:21Four processes from one starting state

So, look at the figure. Small p, the pressure, goes up, and capital V, the volume, goes across. All four lines start from one state, p naught, V naught; the small zero below means the starting value. The flat line is isobaric, constant pressure; straight down is isochoric, constant volume. The curve p V equals constant is isothermal; the steeper one, p V to the power gamma, the Greek letter, is adiabatic; a reversible adiabatic, to be exact. With me? Now, the first law, in the course's sign convention: capital Q is the heat given to the gas, capital W the work done by the gas, and delta U, the little triangle delta, the change in internal energy. So Q equals delta U plus W, okay? For an ideal gas, p V equals n R T: n moles, R the gas constant, capital T the temperature in kelvin. And delta U equals n C v delta T; C v, the C with a small v below, is the molar heat capacity at constant volume. This holds in every process, not only here. Clear?

1:32Isochoric: the volume is fixed

First, isochoric: the volume stays fixed, the line going straight down. Work done by a gas is the area under its p V curve. So with no change in volume, how much work does the gas do? Yes, exactly, the work is zero! No change in volume, no area, so W equals zero; the double arrow means so. Right? So in the first law, all the heat goes into internal energy: Q equals delta U, which is n C v delta T, and nothing leaves as work. Got it? And with V fixed, p V equals n R T makes the pressure grow with temperature. This sign, like a stretched alpha, means proportional to: p is proportional to T, okay?

2:22Isobaric: the pressure is fixed

Next, isobaric: the pressure is fixed, the flat line. The area under it is a rectangle: p times delta V, the change in volume. And by p V equals n R T, that equals n R delta T, as on screen. Clear? Now, the pressure is constant here. So for delta U, do we use C v, or C p, the molar heat capacity at constant pressure? We still use C v, absolutely! Delta U equals n C v delta T in every process, remember? Writing C p here is a classic mistake. Do not fall for it. So C p comes in for the heat. Q equals delta U plus W, n C v delta T plus n R delta T, and since C p equals C v plus R per mole of an ideal gas, Q is n C p delta T, as on screen. With me? And with p fixed, the volume is proportional to the temperature: heat the gas, and it expands. Okay?

3:26Isothermal: the temperature is fixed

Now, isothermal: the temperature stays fixed, so p V stays constant. T does not change, so delta U, which is n C v delta T, is zero. Simple, right? Now, our puzzle. Say the gas expands, but the temperature is fixed. Does it need any heat at all? Yes, it does need heat! With delta U zero, the first law gives Q equals W, okay? Every joule of heat we give goes straight out as work done by the gas. That solves our puzzle! And in an isothermal compression it is the reverse: the work done on the gas leaves as heat. Now the work: W equals n R T times the natural log, written l n, of V two over V one. V one and V two, with small ones and twos below, are the starting and final volumes. Since p V is constant, this ratio also equals p one over p two. Clear?

4:26Polytropic process: pV to the power x is constant

Here is a topper's view of them all as one family: p V to the power x equals constant, a polytropic process. Its work is p one V one minus p two V two, over x minus one, or n R times T one minus T two, over x minus one. Right? Put x equal to zero: p is constant, isobaric. Put x equal to one: p V is constant, isothermal. Put x equal to gamma: a reversible adiabatic. And as x tends to infinity, the sideways eight, it becomes isochoric; this arrow means tends to. With me? But careful. What happens to this work formula when x equals one? The denominator, x minus one, becomes zero, and the formula breaks down! For isothermal work, always use the log formula on screen.

5:23Example 1: 2 mol monatomic gas heated at constant pressure

Got it? Now, example one. Two moles of a monatomic gas are heated at constant pressure, from three hundred to four hundred kelvin; capital K is the kelvin. Find Q, W and delta U, okay? Pause the video and try it first. Done? Okay, let's check. Here n is two and delta T is one hundred kelvin. For a monatomic gas, C v is three over two R and C p is five over two R, with R equal to eight point three one four joules per mole per kelvin; capital J is the joule, m o l the mole, and the power minus one means per. Clear? At constant pressure, the heat is n C p delta T; the dots mean times. Two times five over two R, times one hundred: four thousand one hundred and fifty seven joules. Right? The work done by the gas is n R delta T: two R times one hundred, one thousand six hundred and sixty three joules. Okay? Now delta U, and remember the trap: it is C v, always. Two times three over two R, times one hundred: two thousand four hundred and ninety four joules. Got it? Now check with the first law. Delta U plus W: two thousand four hundred and ninety four, plus one thousand six hundred and sixty three, is four thousand one hundred and fifty seven joules. That is exactly our Q!

6:57Example 2: doubling the volume, isothermal vs isobaric

Now a harder one, example two. From the same state, a gas doubles its volume, in part a isothermally, in part b isobarically. Which case does more work, and absorbs more heat? Take one mole of monatomic gas at three hundred kelvin. Pause the video and try both parts. Done? Okay, let's check. Part a: n is one and V two over V one is two, so Q equals W equals R T times the natural log of two. That is one thousand seven hundred and twenty nine joules. With me? Now part b. The pressure is fixed and the volume doubles. So what happens to the temperature? It doubles too, exactly! Volume is proportional to temperature here, so T goes from three hundred to six hundred kelvin; this arrow means goes to. So delta T is three hundred kelvin. Right? So W equals n R delta T, R times three hundred: two thousand four hundred and ninety four joules. And Q equals n C p delta T, five over two R times three hundred: six thousand two hundred and thirty six joules. Okay? So the isobaric path wins on both. See the figure: the flat isobaric line stays above the isothermal curve, so the area under it, the work, is bigger. And the isobaric gas also heats up, which needs extra heat. Clear?

8:31Quick revision: W, Q and delta U in each process

So, quick revision. One: isochoric, no work, so Q equals delta U; isobaric, Q equals n C p delta T, okay? Two: isothermal, delta U is zero, so Q equals W, n R T times the natural log of V two over V one. Three: delta U equals n C v delta T in every process of an ideal gas. Now go and try the practice problems on this page. All the best!