Mixed Problems: Set 3

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Two moles of helium at three hundred kelvin are connected to one mole of oxygen at six hundred kelvin, in an insulated vessel. What is the final temperature? Weight the temperatures by the moles: two times three hundred, plus six hundred, all over three, gives four hundred kelvin, right? Hold that answer; we check it in example two. This is set three of mixed problems on kinetic theory, in the JEE Main pattern, and most of them come down to counting degrees of freedom. So, let's begin.

0:38The toolkit for this set

First, the toolkit for the whole set. Capital U, the internal energy, is f over two, n R T, where f is the number of degrees of freedom, n the number of moles, R the gas constant, and T the temperature in kelvin. C v, with a small v as subscript, is the molar heat capacity at constant volume, f over two times R, and gamma, the Greek letter, is one plus two over f. With me? So everything depends on f. Let me ask you this: for rigid molecules, with no vibration, what is f for a monatomic gas, a diatomic gas, and a non-linear one? Yes, exactly, three, five and six! Every molecule has three of translation; a diatomic molecule adds two of rotation, and a non-linear one adds three. Okay? And at high temperature, molecules can also vibrate. Each vibrational mode adds two to f, because it stores both kinetic and potential energy. So read carefully: rigid means no vibration. Clear?

1:39Idea 1: energy is shared by degrees of freedom

Now, idea number one. Part a asks what fraction of the internal energy of a diatomic gas at room temperature is rotational. The key is equipartition: each degree of freedom gets the same share, half R T per mole. Okay? So first, at room temperature, what is f for a diatomic gas? It is five, that is correct, because at room temperature there is no vibration. Three of translation plus two of rotation gives five. Right? So rotation holds two of the five equal shares. Two over five is forty percent, written with the percent sign. Clear? Now part b: the heat needed to warm one mole of water vapour by ten kelvin at constant volume; mol means mole, and capital K means kelvin. The question says non-linear and rigid. So what is C v? Water, H two O, where the small two means two hydrogen atoms, is non-linear and rigid, so f is six. So C v is six over two times R, which is three times R, right? At constant volume, the heat Q equals n times C v times delta T, where delta, the little triangle, means the change in temperature. So Q is one, times three, times eight point three one four, times ten. Three times eight point three one four gives twenty four point nine four two. Times ten, that gives about two hundred and forty nine joules, written J; the wavy equals sign means about. Okay?

3:12Idea 2: from rms speed to temperature

Now, idea number two. At what temperature is the root mean square speed, the r m s speed, of nitrogen one kilometre per second? On screen, k m s to the power minus one means kilometres per second. Got it? Start from v r m s, the v with the subscript r m s, which equals the square root of three R T over capital M, where capital M is the molar mass. Square both sides and turn it round, and the double arrow gives T equals capital M times v squared, over three R, okay? Now, nitrogen's molar mass is twenty eight, so we put capital M equals twenty eight, right? Oh no, careful, put twenty eight there and you get over a million kelvin! Capital M must be in kilograms per mole: nitrogen, N two, is twenty eight grams per mole, which is zero point zero two eight kilograms per mole. And the speed must be in metres per second, so one kilometre per second is one thousand metres per second. Clear? Now v squared is one thousand squared, which is ten to the power six. So T equals zero point zero two eight times ten to the power six, all over three times eight point three one four, that is, twenty eight thousand over twenty four point nine four two. Divide, and T is about one thousand one hundred and twenty three kelvin. Got it?

4:29Example 1: degrees of freedom and gamma, oxygen-argon mix

Now, example one. Here a mixture contains two moles of oxygen and three moles of argon. Find the effective number of degrees of freedom, and gamma. Pause the video and try it first. Done? Okay, let's check. Oxygen is diatomic, so f one, with the small one below, is five. Argon is monatomic, so f two, for argon, is three. Right? So the effective f is just five plus three, over two, which is four, isn't it? No, no, wait, that is the trap! There are more moles of argon than oxygen. So we average f by moles: f with the subscript e f f, for effective, equals n one f one plus n two f two, over n one plus n two. Okay? Put the numbers in: two times five is ten, and three times three is nine, so the top is nineteen, and the bottom is five. Nineteen over five gives three point eight. Clear? Now gamma equals one plus two over three point eight. Two over three point eight is about zero point five two six, so gamma is about one point five three, and that's it. Got it?

5:39Example 2: final temperature on mixing helium and oxygen

Now example two, and it is our puzzle from the start. An insulated vessel holding two moles of helium at three hundred kelvin is connected to another holding one mole of oxygen at six hundred kelvin. Find the final temperature. Here is a hint: think about energy, not temperature. Pause the video and try it first. Done? Okay, let's check. Our guess was the mole-average of the temperatures, four hundred kelvin. Is that right? No, it is not, and that is the trap: temperature is not what is conserved, energy is. The vessel is insulated, so the total internal energy stays the same, and each gas has capital U equals f over two, n R T, with me? Helium is monatomic, so three halves; oxygen is diatomic, so five halves. The dot means times. Before mixing: two times three halves R times three hundred, plus one times five halves R times six hundred. After mixing, both gases are at the same final temperature T, so the right side is two times three halves, plus one times five halves, all times R T, right? Now R cancels on both sides. On the left, nine hundred plus one thousand five hundred gives two thousand four hundred. On the right, three plus five halves is five point five. So two thousand four hundred equals five point five times T, okay? So T equals two thousand four hundred over five point five, which gives about four hundred and thirty six kelvin. So it is not four hundred kelvin at all. Each mole of oxygen, the hotter gas, stores five halves R per kelvin, more than the three halves R of a mole of helium, so oxygen pulls the final temperature above the mole-average of four hundred. Puzzle solved: when gases mix, conserve the energy.

7:26Quick revision: equipartition, rms speed, mixing gases

Got it? So, quick revision. Each degree of freedom gets an equal share, half R T per mole, so C v equals f over two times R, okay? For v r m s, the root of three R T over capital M, use capital M in kilograms per mole and T in kelvin. For a mixture, average f by moles, and when gases mix, conserve the internal energy, capital U, not the temperature. Now go and try the practice problem on this page; it uses this same connected-vessel setup. All the best!