Overview

Transcript

0:00Introduction

Hello everyone! Okay, a quick puzzle before we start. Here is a reaction that gives out heat: delta H is minus one hundred and twenty kilojoules. Delta S is minus one hundred and fifty joules per kelvin, and the temperature is four hundred kelvin. One student puts these straight into the Gibbs equation, and gets delta G equal to plus fifty nine thousand, eight hundred and eighty kilojoules! Huge, and positive. Is that believable? Hmm, something is wrong, right? You have finished six chapters of thermodynamics, and now you test yourself with three timed sets of JEE Main style problems. I will show you how to use them, and we will catch this mistake together. So, let's begin.

0:47How to use this chapter

So, what is inside this chapter? Three timed sets, with fifteen problems in each. They are original problems, written in the style of recent JEE Main papers. Okay? Now, one question for you. In the real paper, does a question tell you which chapter it comes from? No, it never does! Spotting the idea is your job. So every set mixes all six chapters of this unit, from the first law right up to Gibbs energy. Just like the real paper. Right? Now the most important instruction. Allow twenty five minutes per set. Sit with a clock, exactly as if it is the real paper, and only then check the solutions. Clear?

1:29The constants for every set

Next, the constants you will use in every set. R, the gas constant: eight point three one four joules per kelvin per mole. J is for joules, K for kelvin, and mol for mole. The small minus ones at the top right simply mean per, okay? And the same R in litre bar units: zero point zero eight three one four litre bar per kelvin per mole. The L here is for litre. Got it? Now wait. One constant, but two different numbers. How can both be correct? Because the units are different, yes! One litre bar is one hundred joules. So zero point zero eight three one four, times one hundred, gives back eight point three one four. Same constant, just different units. With me? Then F, the Faraday constant: ninety six thousand five hundred coulombs per mole, with C for coulombs. You need it in the problems on cells. Right? And remember, use these values unless the problem gives its own. Okay?

2:35Topper's trick: a hundred seconds a problem

Now let me share a small trick that toppers use. Before you start a set, count. Fifteen problems, and twenty five minutes on the clock. So how much time can you give each problem? Twenty five minutes is fifteen hundred seconds. The sign on screen is the division sign. Fifteen hundred over fifteen: one hundred seconds, that's it. So, about one minute and forty seconds for each problem. Read it, spot the idea, write the answer. Clear? Now suppose one problem takes you five minutes, but you get it right. Right answer, so all good? No, no! Those extra minutes were taken from the other problems. So if you are still stuck at one hundred seconds, mark it, move on, and come back at the end. Collect the easy marks first. Got it?

3:24Example 1: delta G at 400 K, the kJ and J trap

Now, remember our puzzle from the start? Here it is, in Set one. The little triangle is the Greek letter delta, meaning change in. So, delta H is minus one hundred and twenty kilojoules, and delta S is minus one hundred and fifty joules per kelvin. Find delta G, the Gibbs energy change, at four hundred kelvin. Pause the video, and try it first. Done? Okay, let's check. The Gibbs equation: delta G equals delta H, minus T delta S, where T is the temperature in kelvin. Clear? So, put the numbers straight in. Minus one twenty, minus four hundred times minus one fifty. That gives plus fifty nine thousand, eight hundred and eighty kilojoules, right? No, no, careful! This is the trap from our puzzle. Delta H is in kilojoules, but delta S is in joules per kelvin. So first, bring them to the same unit. Divide by one thousand: minus one hundred and fifty joules per kelvin is minus zero point one five zero kilojoules per kelvin. Clear? Now substitute. Delta G equals minus one twenty, minus four hundred times minus zero point one five zero. With me? Four hundred times zero point one five zero is sixty, and minus times minus is plus. So, minus one twenty plus sixty: minus sixty kilojoules. That is the answer. Got it? And what does the sign tell us? The sign on screen means less than: delta G is less than zero. At constant temperature T and pressure P, that means the reaction is spontaneous. So our fifty nine thousand mystery was just a units mix-up.

5:08Example 2: final T, adiabatic expansion against 1 bar

Right? Now a harder one, from Set two. One mole of a diatomic ideal gas, at four hundred kelvin and ten bar, expands adiabatically against a constant one bar, until its own pressure is one bar. Find T two, the T with a small two below it: the final temperature. Pause the video, and try it first. Done? Okay, let's check. Adiabatic, so we use the Chapter two formulas, like P V to the power gamma equals constant, right? No, no, careful! Look at the words: against a constant one bar. P with the small e x t is the external pressure, the outside pressure. It stays constant while the gas expands, so this change is irreversible. And P V to the power gamma, with gamma the Greek letter you see on screen, works only for a reversible change. Got it? So, go back to the first law. Adiabatic means no heat goes in or out, so q is zero. Delta U equals q plus w, which is just w. All the work comes out of U, the internal energy. Clear? Now both sides. Left: delta U equals n C V, times T two minus T one. Here n is the number of moles, and C V, the C with a small V below it, is the molar heat capacity at constant volume. Right: the work, minus P ext times the change in volume, with each volume equal to n R T over P, from the ideal gas law. Okay? Now, for a diatomic ideal gas, what is C V? Five over two R, correct! Now put in n equal to one, T one equal to four hundred, P one equal to ten bar, and P two and P ext both one bar. So the right side is minus one, times R T two over one, minus R times four hundred over ten. With me? Every term has R, so cancel it. And four hundred over ten is forty. So, five over two, times T two minus four hundred, equals minus, bracket, T two minus forty. Right? Open the brackets: two point five T two, minus one thousand, equals minus T two, plus forty. Collect the T two terms: three point five T two equals one thousand and forty. The double arrow means which gives, and the wavy equals sign means approximately. So T two is about two hundred and ninety seven kelvin. No heat came in, and the gas did work, so it cooled. Okay?

7:42Quick revision: timing, same units, irreversible adiabatic

So, quick revision. Three timed sets of fifteen problems: twenty five minutes a set, about one hundred seconds a problem. Before you substitute, bring every quantity to the same units: delta H in kilojoules, so delta S in kilojoules per kelvin. And for an adiabatic change against a constant external pressure, use delta U equals w, not P V to the power gamma. Now open Set one, start your clock, and begin. You did the hard work in six chapters, and this is where it pays off. All the best!