Mixed Problems: Set 4

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. In a very good lab vacuum, called an ultra-high vacuum, the gas can have a pressure of just ten to the power minus six pascals. That is almost nothing, right? So how many molecules are left in each cubic centimetre? Is it a handful, or maybe a thousand? Hold your guess; by the end you will know the answer. First, three quick ideas, then two solved examples. Questions like these come up often in JEE Main, so let's begin.

0:32Idea 1: counting molecules

Idea one is about counting molecules. The ideal gas equation, p V equals n R T, with n the number of moles, can also be written p V equals capital N k T, with capital N the number of molecules. Small k is Boltzmann's constant, R over N A, with N subscript capital A for Avogadro's number. It is one point three eight times ten to the power minus twenty three joules per kelvin, the J K to the power minus one on screen. Okay? Now divide both sides by V, the volume. Capital N over V, the number of molecules in each cubic metre, is the number density, n with the small subscript V on screen. So n V equals p over k T, with T in kelvin, always. Right? That gives molecules per cubic metre, but questions often ask per cubic centimetre. So tell me, do we multiply by ten to the power six, or divide? We divide, exactly! One cubic metre, the m cubed, is ten to the power six cubic centimetres, the c m cubed, so each cubic centimetre holds only a millionth of the molecules. Clear?

1:39Idea 2: energy depends on temperature only

Idea two is a problem from this set. At what temperature does a helium atom have the same average translational kinetic energy as an oxygen molecule at three hundred kelvin, the capital K on screen? Got it? First, the result we need. The average translational kinetic energy of one molecule is epsilon, the Greek letter, with a bar on top and the subscript trans. It is one half m, the mass of one molecule, times the average of v squared, shown by the bar, and it equals three halves k T, okay? Now let me ask you something. Helium atoms are much lighter than oxygen molecules. So should the helium be hotter than three hundred kelvin, or colder? It is neither, and that was the trap! Look at three halves k T: there is no mass in it at all. The double arrow means so: the same temperature gives the same energy, for every gas. With me? So the answer is three hundred kelvin, the same as the oxygen. The lighter helium atoms just move faster. Right? Here is one more warning. In most JEE questions, average kinetic energy per molecule means this translational part. If rotation is included, a rigid diatomic molecule has five halves k T, and that leads us to idea three. Clear?

3:02Idea 3: degrees of freedom and gamma

Idea three is equipartition. In thermal equilibrium, each degree of freedom, that is, each independent squared term in a molecule's energy, gets one half k T on average. Okay? Rigid means no vibration. Then f, the number of degrees of freedom, is three for a monatomic gas like helium, five for a diatomic or linear molecule, which adds two rotations, and six for a non-linear molecule like water, with three rotations. Right? From f, everything follows. C v, with the small subscript v, the molar heat capacity at constant volume, is f over two times R, C p at constant pressure is C v plus R, and gamma, the Greek letter, is C p over C v, which is one plus two over f. Got it? Now, a mixture. C v mix, for the mixture, is the sum of n i times C v i, over the sum of n i. The Greek sigma means sum, the subscript i runs over the gases, and n i is the number of moles of each. So it is the average of the C v values, weighted by moles. With me? So here is my question. For gamma of a mixture, can we simply average the gammas of the gases? No, we cannot, and this is a classic trap! Gamma mix is C v mix plus R, over C v mix. Average the C v values first, never the gammas. Clear?

4:26Example 1: molecules per cm3 in a 10^-6 Pa vacuum at 300 K

Now, Example one, which is our puzzle from the start. An ultra-high vacuum has a pressure of ten to the power minus six pascals, the P a on screen, at three hundred kelvin. How many molecules are in each cubic centimetre? Pause the video and try it first. Done? Okay, let's check. By idea one, n V equals p over k T, so that is ten to the power minus six, over one point three eight times ten to the power minus twenty three, times three hundred. Okay? Let us do the bottom first. One point three eight times three hundred gives four hundred and fourteen, so k T is four point one four times ten to the power minus twenty one joules, the capital J on screen. Right? Now divide the top by the bottom. Ten to the power minus six, over four point one four times ten to the power minus twenty one, gives about two point four times ten to the power fourteen per cubic metre, the m to the power minus three; the wavy equals sign means approximately. With me? Last step: for each cubic centimetre, we divide by ten to the power six. Two point four times ten to the power fourteen, over ten to the power six, gives two point four times ten to the power eight. That is two hundred and forty million molecules in every cubic centimetre! So the vacuum is not so empty after all.

5:52Example 2: gamma of 1 mol hydrogen plus 1 mol water vapour

Now the harder one, Example two. Find gamma for a mixture of one mole of hydrogen and one mole of water vapour; the m o l on screen means mole. Both gases are rigid, and the water molecule is non-linear. Here is a hint: find C v of each gas first. Pause the video and try it. Done? Okay, let's check. Hydrogen's gamma is one point four, and water's is about one point three three. So we just take the average of the two, right? No, wait, this is the trap from idea three! We average C v, never gamma. So, let us take hydrogen first. H two is rigid and diatomic, so f equals five, and C v is five halves R, which is two point five R, okay? Water, H two O, is rigid and non-linear, so f equals six, and its C v is six halves R, which is three R, right? Now the mixture, with one mole of each. C v mix is one times two point five R, plus one times three R, over one plus one. Five point five R over two gives two point seven five R, with me? Finally, gamma mix is C v mix plus R, over C v mix, which gives three point seven five R over two point seven five R, and the R cancels. Three point seven five over two point seven five simplifies to fifteen over eleven, about one point three six. Clear?

7:18Quick revision: p over kT, 3/2 kT, mixture gamma

So, quick revision. To count molecules, n V equals p over k T; divide by ten to the power six for each cubic centimetre. Three halves k T per molecule depends on temperature only, not on the gas. And for mixtures, average C v by moles, then gamma is C v plus R, over C v. Now try the practice problem on this page, another mixture. All the best!