Spontaneity and the Second Law

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Put one drop of ink into a glass of water. What happens? It spreads, slowly, through the whole glass. Have you ever seen it gather back into one neat drop? Never, right? But the first law of thermodynamics only asks that energy is conserved, and energy would be conserved in the reverse too. So who decides which way nature goes? Today we meet the answer: entropy, and the second law. Questions on this do come up in JEE, so let's begin.

0:38Spontaneous: by itself, not necessarily fast

So, first, what does spontaneous mean? It means a process that takes place by itself, once it is started. Nobody has to keep pushing it. Okay? Look at some everyday examples. Heat flows from a hot body to a cold body; the arrow on screen means goes to. Gases spread out to fill all the space available. And water runs downhill. Never the other way round by itself. Right? Now tell me. If a process is spontaneous, it must also be fast, isn't it? No, no, careful! Spontaneous does not mean fast. At room temperature, diamond turning into graphite is spontaneous. But it is so slow that nobody can even measure it. Spontaneous is about the direction, not the speed. Clear? And now, back to our ink puzzle. Energy is conserved whether the ink spreads or gathers back. The first law allows both directions, so it cannot decide which way a process goes. We need something new. With me?

1:43Entropy: a measure of dispersal

That something new is entropy, written with the capital letter S on screen. Entropy measures how spread out, or dispersed, energy and matter are. In simple words, disorder. More spread out means more entropy. Got it? And entropy is a state function: its value depends only on the state, not on the path. But heat, the small q, is not a state function. It depends on the path. Okay? Now, we want the change in entropy from the heat. But the heat depends on the path. So which path should we use? The reversible path, yes! For a reversible transfer of heat at temperature T, delta S equals q rev over T, heat divided by temperature. The little triangle is the capital Greek letter delta, meaning change in. The q with a small rev written below is the heat of the reversible path, and T is the temperature in kelvin. The unit in brackets, J K to the power minus one, is read as joules per kelvin. Right?

2:49The second law

Now, the big law. Entropy can change in two places: the system, and its surroundings. So delta S total equals delta S sys plus delta S surr. The small words written below, total, sys and surr, are short for total, system and surroundings. Clear? So here is my question. In a spontaneous process, what happens to this total entropy? It increases, exactly! This is the second law of thermodynamics. In any spontaneous process, the entropy of the universe increases: delta S total is greater than zero. And our ink puzzle is solved: spreading increases the entropy, and gathering back would decrease it. Perfect! And at equilibrium, delta S total equals zero. No increase, and no decrease. Okay? Now one condition. For an isolated system, which exchanges neither energy nor matter with the outside, delta S is greater than or equal to zero. The line under the greater than sign means or equal to. Zero for a reversible process, and positive for an irreversible one. With me?

3:58Trap: endothermic, yet spontaneous

Now, a famous trap. Ammonium nitrate dissolves in water all by itself. But this dissolving is endothermic: it takes in heat, and that lowers the entropy of the surroundings. So what is driving it? The system itself, yes! When the solid dissolves, its particles spread out through the water, so the entropy of the system goes up. And it goes up by more than the surroundings lose. So the total still increases. Got it? So never judge by the system alone. Only delta S total greater than zero matters. Delta S sys alone can even be negative, and the process can still be spontaneous, as long as the surroundings gain more. Right?

4:42Two more ways to state the second law

Now, the second law has two more equivalent statements. First, Kelvin and Planck: no cyclic engine, that is, one working in a cycle, can turn heat taken from one reservoir completely into work. Notice, in a cycle: that is what an engine does, again and again. Okay? Second, Clausius: heat cannot flow by itself from a colder body to a hotter body. Notice the words, by itself. That is the key. Clear? And this explains your car. Its engine takes heat from a hot body, but it cannot turn all of it into work. Some heat must be rejected to a colder body, and the exhaust does exactly this. Right?

5:22Example 1: which of these processes are spontaneous?

Now, Example one. Which of these are spontaneous? Ice melting at twenty five degrees Celsius; the small circle means degrees. Water freezing at twenty five degrees Celsius. Gas filling an evacuated flask, that is, a flask with all the air pumped out. And heat passing from a cold body to a hot body. Pause the video, and try it. Done? Okay, let's check. At twenty five degrees Celsius, ice melts all by itself: spontaneous. Water freezing at that temperature is the reverse, so it never happens by itself: not spontaneous. Okay? Gas filling an evacuated flask: gases spread out, so yes, spontaneous. Right? Now the last one, and be careful. When heat passes from the cold body to the hot body, energy is still conserved. So the first law allows it, and it must be spontaneous, isn't it? No, no, wait! The first law cannot decide the direction, remember? Clausius tells us heat cannot flow by itself from a colder body to a hotter one. So, not spontaneous. Clear? So the answer is the first and the third: ice melting, and the gas filling the flask. That's it!

6:40Example 2: total entropy change, heat from 400 K to 300 K

Now a numerical one. One thousand joules of heat flows from a body at four hundred kelvin to a body at three hundred kelvin. The capital J on screen stands for joules, and the capital K for kelvin. Find the total entropy change. Pause and try. Hint: heat over temperature, for each body. Done? Okay, first a quick thought. The hot body loses one thousand joules, and the cold body gains the same one thousand joules. So the two entropy changes cancel, and the total is zero, right? No, no, careful! That is the trap. We divide the heat by the temperature, and the two temperatures are different. The hot body loses heat, so its change is negative. Delta S hot, with the small word hot written below, is minus one thousand over four hundred: minus two point five joules per kelvin. Got it? Now, the cold body gains the same heat, so its change is positive. Delta S cold is plus one thousand over three hundred, about plus three point three three joules per kelvin; the wavy equals sign means approximately. See, the same heat at a lower temperature gives a bigger entropy change. With me? So delta S total is minus two point five, plus three point three three: plus zero point eight three joules per kelvin. Greater than zero, so by the second law, this heat flow is spontaneous. And the reverse flow would make the total negative, so it never happens on its own. Perfect!

8:13Quick revision: spontaneity, entropy and the second law

So, quick revision. One: spontaneous means it takes place by itself, not that it is fast. Two: delta S equals q rev over T, and in a spontaneous process, delta S total is greater than zero. Three: judge by the total entropy, never by the system alone. Now go and try the practice problems on this page. All the best!