The Kinetic Interpretation of Temperature

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Hydrogen and oxygen are mixed in one box, at one temperature. An oxygen molecule is sixteen times heavier than a hydrogen molecule. So, on average, which molecule carries more kinetic energy? Some of you say oxygen, because it is heavier. Others say hydrogen, because it moves faster. Hold your answer; one short formula will settle it. This idea comes up often in JEE, so let's begin.

0:33From pressure to temperature

So, we start from the last lesson. Kinetic theory gave us p V equals two thirds of E trans, where p is the pressure and V is the volume, okay? E with the subscript trans is the total translational kinetic energy of all the molecules. Right? Now, the ideal-gas equation written with molecules: p V equals N k T, okay? Capital N is the number of molecules, T is the temperature in kelvin, and small k is Boltzmann's constant, one point three eight times ten to the power minus twenty three joules per kelvin, written J K to the power minus one. Clear? Now look at these two lines together. Both have the same left side, p V, right? So what can we say about their right sides? Yes, exactly, the right sides must be equal! So two thirds of E trans equals N k T, and the double arrow on screen means that the next result follows: E trans equals three halves of N k T, got it? Now, this energy is shared among all N molecules. So, on average, how much does one molecule get? Divide by capital N, and one molecule gets three halves k T, correct! The Greek letter epsilon with a bar on top is the average energy of one molecule, again with the subscript trans. It also equals half m times v squared bar, where m is the mass of one molecule, and the bar over v squared means the average of v squared. Okay? And what about a whole sample of n moles? Small n is the number of moles and R is the gas constant, and N times k equals n times R, as the brackets show. So E trans equals three halves n R T, for n moles. With me?

2:18Temperature measures molecular energy

Now look closely at this result: three halves k T, and nothing else. Let me ask you this: does it depend on the mass of the molecule, or on which gas it is? No, it does not, and that is the beautiful part! Small k is a constant, so only the temperature can change this energy. It does not depend on the kind of gas, on its pressure p, or on its volume V, okay? That is the kinetic meaning of temperature. Clear? So, back to our puzzle. Hydrogen and oxygen in one box, at one temperature. Which molecule has more translational kinetic energy on average? The answer is neither, because they are exactly equal! In a mixture, every component has the same three halves k T per molecule, since they share one temperature. The heavy oxygen molecule simply moves slower, and the light hydrogen molecule faster. So our puzzle is solved, very good!

3:18Trap: which kinetic energy?

Now, one trap before the examples. Suppose a question asks for the average kinetic energy per molecule of oxygen gas. Is the answer always three halves k T? In most JEE questions, yes. Average kinetic energy per molecule usually means the translational energy, three halves k T, the same for every gas at a given temperature. Okay? But careful, an oxygen molecule can also rotate. If rotation is included, a diatomic molecule, made of two atoms like oxygen or nitrogen, has five halves k T, not three halves. So read the question: translational only, or rotation included? Got it?

4:03Example 1: average kinetic energy at 27°C in J and eV

Time for example one. Find the average translational kinetic energy of a gas molecule at twenty seven degrees Celsius, shown as a small circle and a capital C, in joules and in electronvolts. Pause the video and try it first. Done? Okay, let's check. The formula is three halves k T, and the temperature is twenty seven, so we put T equals twenty seven, right? No, no, careful, that is the trap! In kinetic theory, temperatures are always in kelvin. So T equals twenty seven plus two hundred and seventy three, which gives three hundred kelvin, written with a capital K, okay? Now substitute: three halves, times one point three eight times ten to the power minus twenty three, times three hundred. Right? Three halves of three hundred gives four hundred and fifty, and four hundred and fifty times one point three eight gives six hundred and twenty one. So the energy is six point two one times ten to the power minus twenty one joules, the capital J on screen. With me? Now electronvolts, written e V on screen. One electronvolt is one point six times ten to the power minus nineteen joules. So divide our answer by this, and we get about zero point zero three nine electronvolts; the wavy equals sign means approximately, and that completes example one.

5:29Example 2: rms speed of nitrogen at 300 K

Now example two, one step harder. Find v r m s for nitrogen at three hundred kelvin. On screen, v with the subscript r m s is the root mean square speed: the square root of the average of v squared. Pause the video and try it. Here is a hint: start from half m times v squared bar. Done? Okay, let's check. Set half m times v squared bar equal to three halves k T, okay? So v squared bar equals three k T over m, and v r m s is its square root, the root of three k T over m; the long tick on screen is the square root sign. Clear? But we usually know R and the molar mass, not k and m. So multiply the top and the bottom of k over m by N A, Avogadro's number, written N with a subscript capital A, okay? Small k times N A is R, and the mass of one molecule times N A is capital M, the molar mass. So v r m s equals the root of three R T over capital M, right? Nitrogen gas, written N two, has a molar mass of twenty eight grams per mole. So we put capital M equals twenty eight, right? Oh no, careful, this is the second trap! R is eight point three one four joules per mole per kelvin, and joules are built from kilograms. So capital M must be in kilograms per mole: twenty eight grams per mole, g mol to the power minus one, is zero point zero two eight kilograms per mole, k g mol to the power minus one. Got it? Now substitute: three, times eight point three one four, times three hundred, over zero point zero two eight, all under the root. Work it out, and v r m s is about five hundred and seventeen metres per second, written m s to the power minus one. Very good, that is example two done!

7:24Quick revision: 3/2 kT for every gas, 5/2 kT trap, units

So, quick revision. One molecule has average translational kinetic energy half m times v squared bar, which equals three halves k T, and n moles have three halves n R T, okay? It depends only on the temperature, so it is the same for every gas, even in a mixture. With rotation included, a diatomic molecule has five halves k T instead. And always put T in kelvin, and capital M in kilograms per mole. Now go and try the two practice problems on this page. All the best!