# Thermodynamics for Dummies: Entropy and temperature

I do not study physics and I have never had a course in thermodynamics. I have no idea what it is about, but I am currently taking a course where we had something about entropy.

Would be great if you could help me in easy words.. Wikipedia and other Websites did not help.

So: We are looking at a box full of gas with N particles in it (see Maxwell's demon). The question is now to look at two different states. The first state ist when the whole box is used for the gas. The second state is when only one half is used.

Now we consider the Transition.

1) -> 2): The second state has less states, hence less entropy. So we lose entropy. This Forces us to do work, to invest energy, otherwise this would not happen (or only with small probability), as the 2nd law of thermodynamics states. So what does this mean for the temperature in this box? Are we considering the temperature in this box or around this box, or both? Does it get warmer or colder? Where?

2) -> 1): This is more probable, and energy is set free. We have more entropy then. But also, this has to do something with temperature, so what are the consequences?

I know how to calculate the energy that has to be invested to enforce 1)->2), but what I do not know is the consequences for the temperatures. So we talk about $\Delta S \geq - N k_B ln 2$ (why $\geq$ and not $=$), the difference in entropy, so the temperature difference is $\Delta Q = T \Delta S$..

Sometimes I have read that more entropy means hotter. As if particles are hotter, they move faster, so there's more uncertainty. But then, I also read that if we go 2) -> 1) that the temperature gets colder. But the entropy grows. So I have no idea. My guess is, that sometimes one talks about the temperature in the gas, and sometimes one talks about the temperature outside this box.

Thank you for your help. Hope I am right here.

• in thermodynamics (which i recommend to study a little more), has many types of transitions e.g: adiabatic, isothermal (same temperatue), isochoric (same volume), etc.. Where sth changes with respect to sth that stays constant Aug 22, 2014 at 19:29
• btw $\Delta Q = T \Delta S$ is not temperature difference, it is energy difference (i.e heat). Heat is not temperature Aug 22, 2014 at 19:30
• This is a big, but very good question. It will require a tour-de-force answer, or a lot of little answers. I have one of the latter: 2 --> 1 does not necessarily mean that temperature decreases. For an ideal gas, the temperature will remain the same. There's no reason the energy per particle should change. But the entropy will increase. For a real gas the temperature may decline, but that would be due to the complicating factor of inter-molecular forces. Considering such things will not help answer your basic questions. Aug 22, 2014 at 21:54
• Oh ... and you say that you know how to calculate the energy to go 1 --> 2. Perhaps, but if you do you must specify what means you are using to get from 1 to 2. Doing it slowly takes a different amount of energy than doing it quickly. Aug 22, 2014 at 21:59

The Second Law of Thermodynamics states that the entropy of the universe always increases or stays constant. This means that we can reduce the entropy of the gas in the box (gas compressed from the whole box to half the box at constant temperature), only if we increase the entropy somewhere else. For example, we can compress the gas, doing work on it, and thus heating it. Then we have to allow the gas to lose this heat to the environment so that it ends at the original temperature, in half the box. The result is that by heating the environment I have increased the entropy of the environment by at least as much as I have decreased the entropy of the gas in the box.

Since most systems are not isolated (they can exchange energy with the environment), I always have to include both the system (box of gas etc) and the environment.

2) -> 1): This is more probable, and energy is set free.

If we allow the gas to expand suddenly into the whole box by removing the partition in the middle, the entropy increases, but there is no change in the energy. This is an irreversible process - entropy increases.

ΔQ=TΔS

This is not generally true. It is only correct when the change is reversible - in other words the change occurs slowly through a series of equilibrium states. You also have to interpret it correctly. ΔQ is the heat added to the system from the outside. ΔS is the increase in entropy of the system. If the process of taking this heat from the outside world is also reversible, the outside world loses ΔQ of heat, and loses ΔS in entropy, so the entropy of the universe is unchanged.

The case above (sudden expansion of the gas) is an example where this equation does not hold. There is no transfer of heat, but the entropy of the gas increases.

To summarise. The temperature of a gas is a fairly easy concept - it is proportional to the average random kinetic energy of the molecules. The entropy is more complicated. Entropy is increased by allowing the gas to occupy more volume, because the molecules can be arranged in space in a larger number of ways (more "states" as you describe it). Entropy is also increased by raising the temperature of the gas, because this gives more ways for the kinetic energy to be distributed.

When does a change in a system occur spontaneously (for example a chemical reaction, or the expansion of a gas)? It occurs either because the change increases the entropy of the system, or because it decreases the total energy of the system. The spare energy leaves the system, heating and thus increasing the entropy of the environment.

Of course, if the course you are taking is not in physics or physical chemistry, it is likely to include a great deal of nonsense when it comes to entropy. The subject can only be understood by disentangling the concepts as I have tried to do.