Set 3

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Take an insulated vessel. One half holds an ideal gas, and the other half is a vacuum, completely empty. Open the gap, and the gas rushes out to fill the whole vessel. No heat comes in, no work is done, and its energy does not change at all. Then why does it happen? And why does it never go back into one half? Hmm, the first law cannot answer this, right? This is Set three of our JEE Main style mixed problems on thermodynamics, and we will settle this puzzle on the way. So, let's begin.

0:35Tool 1: the first law, with its signs

So, first tool: the first law. Delta U equals q plus w. The little triangle is the Greek letter delta, meaning change in, and U is the internal energy. q is heat, w is work, and both count as positive when energy goes into the system. Okay? One from our set. In one cycle, a gas absorbs four hundred joules of heat, J for joules, and gives out one hundred and fifty. Find the net work done by it. But first, after one full cycle, what is delta U? Zero, yes! The gas comes back to the same state, so its internal energy is back where it started. Delta U is zero, so q equals minus w. Right? The net heat in is four hundred minus one fifty: two hundred and fifty joules. So w is minus two hundred and fifty joules. Minus means the gas does the work: two hundred and fifty joules of it. Clear? Now, back to our puzzle, also in our set. This is called a free expansion. P ext, the P with a small e x t below, is the external pressure. Into a vacuum it is zero, so w is zero. And the vessel is insulated, so q is zero. With me? So delta U, q plus w, is zero. For an ideal gas, the energy depends only on temperature, so delta T, the change in temperature, is zero too, and so is delta H, the enthalpy change. Everything is zero! So what still changes? The entropy, exactly! Delta S, the entropy change, is greater than zero; that sign on screen means greater than. The gas spreads out, and in an isolated system, a spontaneous change always raises the entropy. That drives it, and that is why it never goes back. Puzzle solved!

2:23Tool 2: enthalpy from bond enthalpies

Now, the second tool. Delta r H, the reaction enthalpy, with a small r below, equals the bond enthalpies of the bonds broken, minus those of the bonds formed. The big Greek letter sigma means sum of. This works for gases. Okay? From our set: estimate delta f H, the enthalpy of formation, with a small f below, for H B r gas; the g in brackets means gas. Each dash is a bond. The bond enthalpies are four thirty six for H H, one ninety three for B r B r, and three sixty six for H B r, in kilojoules per mole. The minus one up top means per. Got it? So, the quick way: break one H H bond and one B r B r bond, and form two H B r. Four thirty six plus one ninety three, minus two times three sixty six. Correct? Oh no, careful! That equation makes two moles of H B r. Formation means exactly one mole of the compound, from its elements. So, half a mole of H two gas and half a mole of B r two gas, and the arrow means give: one mole of H B r. One small note: the reference state of bromine is actually the liquid, but the question gives the B r B r value for bromine gas, so here we start from B r two gas. Right? So, one half times four thirty six plus one ninety three, minus three sixty six. That is minus fifty one point five kilojoules per mole. Negative, so forming H B r gives out heat. Clear?

3:57Example 1: melting point from delta H and delta S

Alright, example one, an easy one. This solid has delta fus H equal to eight point zero kilojoules per mole; fus, written small below, means fusion, that is, melting. And delta fus S is twenty five joules per kelvin per mole, with K for kelvin. Find its melting point. Pause the video, and try it first. Done? Okay, let's check. At the melting point, the solid and the liquid are in equilibrium. And at equilibrium, at constant temperature and pressure, delta G, the Gibbs energy change, is zero. Okay? Delta G is delta H minus T delta S, where T is the temperature. Put it equal to zero. The double arrow means which gives: T equals delta H over delta S, simple! On the graph, that is where a line crosses zero. For melting, delta H and delta S are both positive, so above that temperature, the solid melts by itself. So, eight point zero over twenty five: zero point three two kelvin, right? No, no, careful! That is almost absolute zero; no ordinary solid melts there. Delta H is in kilojoules, but delta S is in joules per kelvin. So first, change delta H to joules: eight thousand joules per mole. Got it? Now, eight thousand over twenty five: three hundred and twenty kelvin. That is the melting point. Same units first, always. Right?

5:26Example 2: work in a two-step isothermal expansion

Now the hard one from our set. One mole of an ideal gas at three hundred kelvin expands isothermally, that is, at constant temperature, from three bar to one bar. But in two steps: first against two bar, then against one bar. Find the total work. Pause and try. Hint: find the volumes first. Done? Okay, let's check. On the right, pressure P goes up, volume V goes across, and each shaded area is the work. Isothermal, so we use the formula under the third graph: n R T, l n, the natural log, of V two over V one, right? No, no, careful! The third graph is the reversible path, with infinitely many tiny steps. Here P ext is held at two bar, then at one bar. Constant P ext in each step means irreversible. So in each step, the work is minus P ext times the change in volume, like the middle graph. Okay? So, the volumes. V equals n R T over P, with n the number of moles and R the gas constant in litre bar units. At three, two and one bar: eight point three one, twelve point four seven, and twenty four point nine four litres; L is for litres. With me? Step one, against two bar, from eight point three one to twelve point four seven litres. w one, the w with a small one below, is minus two times their difference: minus two times four point one six. Right? Now step two, against one bar. From which volume? From twelve point four seven, yes, where step one ended, not from the very start! Up to twenty four point nine four. So w two is minus one times twelve point four seven: minus twelve point four seven. Clear? Now add them. Minus eight point three two, minus twelve point four seven: minus twenty point seven nine litre bar. Got it? One litre bar is one hundred joules, so that is minus two thousand and seventy nine joules. The wavy equals sign means approximately: about minus two point zero eight kilojoules. Minus, because the gas does the work. And the middle graph shows it: two steps, more work than one. Okay?

7:44Quick revision: first law, bond enthalpies, melting point

So, quick revision. First law: delta U equals q plus w. In a cycle, delta U is zero; in a free expansion, q, w and delta U are all zero, but the entropy rises. Bond enthalpies: broken minus formed, for one mole of product. Melting point: delta H over delta S, in the same units. And in steps, add minus P ext delta V for each. Now go and try the practice problems on this page. You can do this. All the best!