Entropy and the Otto Cycle

Transcript

0:00Introduction

Hello everyone! First, a small puzzle. Leave hot tea in a cool room, and heat flows out of the tea, into the room. Now imagine the opposite: the room gives heat back, and the tea starts boiling, all by itself. Energy is still conserved, so the first law would allow it. And yet, it never happens. So what decides the direction? Today, a new quantity called entropy answers this, and then comes the Otto cycle of a petrol engine. Both of these can come up in JEE, so let's begin.

0:33Entropy: heat divided by temperature

So, the new quantity is entropy, written as capital S, and for a reversible process, d S equals d Q over T, okay? The upright d means a tiny amount: d S is a tiny change in entropy, and d Q is a tiny heat given to the system at temperature T, so entropy change is heat over temperature. Sign convention, the same as in the first law. d Q is the heat given to the system: positive when it goes in, negative when it comes out, as the greater than and less than signs show. And T is always in kelvin. Right? Now think. Give the same small heat to a hot body, and to a cold body. Where does the entropy change more? In the cold body, exactly! T sits in the denominator, so a smaller T gives a bigger change. This will solve our tea puzzle soon. Clear? Next, like internal energy, entropy is a state function. So delta S, where the capital Greek delta means change in, depends only on the end states, not on the path. Got it?

1:39Entropy change in four standard processes

Now, the standard processes. Reversible isothermal first: T stays fixed, so delta S is simply Q over T, with me? For an ideal gas at constant temperature, this becomes n R times the natural log of V two over V one. Here n is the number of moles, R the gas constant, the l n is the natural log, and V one and V two, with a small one and two below, the starting and final volumes. Expansion gives a positive log: entropy goes up. Okay? At constant pressure, the small p, delta S equals n C p times the natural log of T two over T one. C p, capital C with a small p below, is the molar heat capacity at constant pressure, and the temperatures are in kelvin. Right? Now, a reversible adiabatic process: d Q is zero at every single step. So what is delta S? Zero, correct! Zero at every step adds up to zero. But notice that word, reversible, okay?

2:43The free expansion trap

Now a classic JEE trap: free expansion, a gas rushing into a vacuum. No heat flows in, so Q is zero; there is nothing to push against, so the work W is zero. So by the first law, delta U, the change in internal energy, is zero too. Clear? No heat flowed at all. So delta S must be zero, right? No, no, careful here! d S equals d Q over T holds only for a reversible process, and free expansion is irreversible. So use the state function idea. Delta U is zero, so for an ideal gas the temperature does not change, and the end states match an isothermal expansion. With me? And entropy depends only on the end states. So delta S equals n R times the natural log of V two over V one, which is positive, as the greater than sign shows. Entropy went up, with no heat at all. That is the trap. Okay?

3:41The second law in terms of entropy

Now the big law. For an isolated system, delta S is greater than or equal to zero; that is the greater than sign with a line below. Equal only for a reversible process; an irreversible one, like free expansion, makes it go up. Right? Back to our tea. Heat leaves the hot tea and enters the cooler room. The tea loses entropy, and the room gains it. Which change is bigger? The room's gain, yes! Same heat, lower temperature, so a bigger heat over temperature. So heat flowing from hot to cold always increases the total entropy, and the reverse would lower it. The second law forbids that, and that solves our tea puzzle. Got it?

4:24The Otto cycle: the petrol engine

Now the petrol engine. Its cycle, the Otto cycle, has two adiabats, with no heat exchanged, and two isochors, where the volume stays fixed. Okay? The key number is the compression ratio, small r: largest volume over smallest volume. Clear? And the efficiency, the Greek letter eta, is one minus r to the power one minus gamma. Gamma, another Greek letter, is C p over C v, where C v, with a small v below, is the molar heat capacity at constant volume. Look at the power: one minus gamma, not gamma minus one. With me? Think like a topper. If r gets bigger, does the efficiency go up, or down? Up, correct! Gamma is more than one, so the power one minus gamma is negative, and r to that power shrinks as r grows. One minus a smaller number gives a higher efficiency. Got it?

5:19Example 1: entropy change when heat flows from hot to cold

Alright, the first example: our tea puzzle with numbers. One hundred joules of heat flows from a reservoir at four hundred kelvin to one at three hundred kelvin; capital J is joules, capital K is kelvin. Find the total entropy change. Pause the video and try it first. Done? Okay, let's check. One hundred joules out, the same one hundred joules in. So the two changes cancel, and the total is zero, right? Oh no, careful here, the two temperatures are different! Each reservoir stays at a fixed temperature, so delta S is Q over its temperature. The hot one, delta S with a small word hot below, gives out heat, so its Q is negative: minus one hundred over four hundred, which is minus zero point two five. The unit, J K to the power minus one, means joules per kelvin, okay? The cold one, with the small word cold below, takes in heat: one hundred over three hundred, about zero point three three three; the wavy equals sign means approximately. Clear? Total: minus zero point two five, plus zero point three three three, which is plus zero point zero eight three joules per kelvin, greater than zero. Positive, just as the second law demands for this isolated pair. The cold side divides by a smaller temperature, so it gains more than the hot side loses. Very good, that is the answer.

6:48Example 2: Otto engine efficiency with r = 8, gamma 1.4

Now a slightly harder one. Find the efficiency of an Otto engine with compression ratio eight, and gamma equal to one point four. Pause the video and try it first. Done? Okay, let's check. Some students write eta equals one minus eight to the power zero point four. Is that right? No, no, careful, that is exactly the trap here! Eight to the power zero point four is more than one, so the efficiency would be negative, which is impossible. The power is one minus gamma, which is minus zero point four, so eta equals one minus eight to the power minus zero point four. With me? Eight is two cubed, so eight to the power zero point four is two to the power one point two, about two point three zero. One over that is about zero point four three five. Okay? So eta is one minus zero point four three five, which is zero point five six five, or fifty six point five percent. Less than one, as it must be, so that is our answer.

7:49Quick revision: entropy changes, second law, Otto cycle

So, quick revision. d S equals d Q over T for a reversible process, with T in kelvin, and entropy is a state function. Reversible isothermal: Q over T; reversible adiabatic: zero; free expansion: n R times the natural log of V two over V one. For an isolated system, delta S is never negative, so heat flows from hot to cold. And the Otto cycle: eta equals one minus r to the power one minus gamma. Now go and try the practice problems on this page. All the best!