Thermodynamics basics.

Urmi Roy

Hi!

I just need some basic things cleared and I would really appreciate if someone helped me out.

Here are a few questions that put forward my doubts:

1. Thermodynamic equilibrium requires mechanical equilibrium--please explain.

2.How many state variables (Voume,pressure,temperature etc.) are needed to specify a thermodynamic system?

3. In order for Boyle's law to be applicable to a process,does the process need to be quasi static and (or) isothermal?(Referring to this,the main thing is--is an isothermal process always quasi static?)

4.What kind of factors does continuum volume (the minimum volume needed to obtain a continuum in the system) depend upon?


5. why is density a thermodynamic quantity?(It says so in my book).

6. What is the significance of classifying systems into 'open','closed' and 'isolated' systems?(I mean in real life,do we consider them while performing adiabatic/isothermal/isochoric processes?)

7. Lastly,what is the difference between the Helmoltz equation and gibb's equation(just basic idea required).

Thanks.
(Btw,there are a lot of questions,but I guess they're short ones!!)

mikelepore

Notice that the ideal gas law PV=nRT can also be written in the form PM=dRT, where M is molar mass, d is density. Derivation: PV=nRT, substitute n=m/M (number of moles = sample mass / molar mass), PV=(m/M)RT, rewrite as PM=(m/V)RT, density d = m/V, PM=dRT.

Urmi Roy

Does that mean that density is a thermodynamic quantity because the pressure or temperature changes cause an overall density change of the gas?

mikelepore

I would say yes to your question. Volume is a thermodynamic variable, and when the volume changes the density changes, therefore density is a thermodynamic variable also.

Urmi Roy

Thanks!
Could you answer some of my other questions too,please?

mikelepore

question 3: Boyle's law just requires that you are comparing two instants in time when the temperature T and the amount of substance n are the same in both cases: P1 V1 = nRT = some constant, P2 V2 = nRT = the same constant, P1 V1 = P2 V2. There is no dependency on the kind of process. The main problems that cause Boyle's Law to fail are when the pressure or density get too high (because electrical intermolecular forces may become significant) or when the temperature gets too low (because a real gas can liquify). Otherwise, the process doesn't affect the correctness of the equation.

mikelepore

Your other questions, I'm afraid I might answer them incorrectly. There may be someone else here who has more certainty.

my_wan

Consider a pressurized tank of air (gas). Now this tank of gas is the same temperature as outside air, so if it was in equilibrium it wouldn't leak under pressure, unless it's not only in thermal equilibrium, but also in mechanical equilibrium. At a fundamental level there's not much difference. Open the valve till all the air leaks out, then heat it and more air leaks out under pressure.

The pressurized tank is not in equilibrium because the gas is denser, due to more mechanical impacts of the molecules. Once in mechanical equilibrium, then heating the tank adds velocity to the gas particles. Now the gas molecules are not in equilibrium, not because there is more, but because they are hitting harder and faster than the molecules outside.

That depends on exactly what you want to specify. Pressure, volume, and temperature is enough for an ideal gas. Temperature is basically the average kinetic energy, 1/2Mv^2, of the molecules. Changing density also changes volume, and temperature due to the kinetic energy of the particles being closer together, thus more energy per volume.

Yes, the temperature is what's defined to stay static in Boyle's law, but the pressure and volume are inversely related. So Boyle's law does not require constant pressure and volume, only temperature, but the ratio of pressure and volume will stay constant though.

This is a more difficult question. It really doesn't make much sense to talk about the temperature of a single particle, though you can define its kinetic energy. The volume doesn't really matter either, but the fewer particles there are the bigger the volume you need to average over to make sense. So primarily it depends on an area big enough so that averaging the kinetic energy of the molecules in it will give you a true average. It's like asking what the average shoe size is. You can't learn that from 1 shoe.

For the same reason I described in question 1. If you compress more gas closer together, then the average kinetic energy per unit volume increases, because the kinetic energy of the molecules are closer together. Increasing temperature also increases the average kinetic energy per unit volume, not by more particles with kinetic energy closer together, but by adding more kinetic energy per particle.

An open and isolated system are opposites. A closed system is a bit different. An isolated system cannot exchange any parts or energy with anything outside that system. A heated gas that can't interact with anything outside that gas can never heat up or cool down, if the volume stays the same. Insulation is an attempt to isolate the air system in a house. It's never perfect under any circumstances. An open system is the opposite and allows parts and energy in and out of the system.

Now a closed system is when no parts are allowed in or out of the system, but energy is. The freon in an air conditioner is 'enclosed', but is what is used to transfer heat out of your house. That's what the big coils outside the house is for, as a heat exchanger.

The is very rough, but basically the Helmholtz equation separates two variables to each side of an equation such that both side are equal to some constant. This allows you to separate those variables into two separate equations, both equal to that constant. It's useful for defining angular frequency in linear trig functions, Fourier transforms, etc., depending on which form you use. I assume, when you say Gibbs equation you mean the Gibbs–Helmholtz equation. It simply defines the change in Gibbs energy as temperature changes.

These last concepts need a better understanding of the previous questions. It would be best to learn how and why the equation of state of an ideal gas are manipulated the way they are. With 3 variables you can hold one of them constant to see how the other two work together. Then choose another variable to hold constant and learn how another pair of variables work together. When you learn how any combination of two variables work together, you can start treating all 3 as variables. That's how you learn about the answers to these questions you are asking.

johng23

The difference in Helmholtz and Gibbs energy:

In general, the equilibrium state of a system is the state of maximum entropy. Since entropy is not a quantity that can be easily understood or measured in real situations, we can define other thermodynamic potentials that are relevant to the conditions at hand. Gibbs energy and Helmholtz energy are two such potentials. You can show that, in the right conditons, the entropy maximum condition for equilibrium is equivalent to either the minimum Gibbs free energy or the minimum Helmholtz free energy.

In chemical reactions that occur commonly on earth, the pressure is fixed at atmospheric pressure, and the temperature is fixed by the surroundings. Then P and T are the controlled variables. In this situation, you can show that maximum entropy is equivalent to minimum Gibbs free energy. Thus, it is the Gibbs energy that is generally used to analyze the system.

In a pressure cooker, the situation is different. In this case, the volume is fixed by the container. Temperature and volume are the controlled variables, and the system will spontaneously evolve until the Helmholtz energy is minimized.

In general, it's all about maximizing entropy, but how that manifests itself depends on the physical constraints of the system.

Urmi Roy

This is exactly I was cofused about initially. It seems that due to the fact that both pressure (hence mechanical equilibrium) and temperature (hence thermal equilibrium) both vary similarly on heating or cooling,there isn't much difference between mechanical equilibrium and thermal equilibrium . I read up on this later,and I found that pressure (hence
mechanical equilibrium ) can be affected in many ither ways also,hence it is different from thermal equilibrium.


Thanks for the clarification! I also figured out that the process to which Boyle's law is applied has to be quasi-static also,else we can't trace the progress of the process and plot it,right?


Thanks,I found your explanation very helpful!



I now understand exactly what these terms mean. Just going a bit further,would it be right to say that for most of our experiments on adiabatic processes,isobaric processes and isochoric processes are conducted with isolated systems? On the other hand,for isothermal process,we need a closed system.


The other answers were also very helpful. Thanks,my_wan!

Urmi Roy

Right,so that means both these equations are used to check the feasibilty of a process without directly taking recourse to entropy,but in two different kinds of situation (Gibb's free energy for P-T invariant processes and V-T invariant processes use Helmoltz free energy).

Thanks,johng23!

Andy Resnick

Not always- a freely-falling body is not in mechanical equilibrium, but can have a well-defined temperature (thermodynamic equilibrium). Two objects may be at mechanical equilibrium (say, at rest toughing each other) but not in thermal equilbirium (different temperatures= heat flow). Heat is not a mechanical phenomenon.

The state space has 5 dimensions- volume, pressure, temperature, entropy and enthalpy. That the state space has an odd number of dimensions, whereas the mechanical state space has an even number, is significant.

Boyle's law is an approximate law which gives accurate predictions for dilute cold gases. AFAIK, the process must simply be isothermal, and does not have to be quasi-static.

The spacing between atoms (mean free path). The volume element must be much larger than this distance.

density of what? mass? energy?

The significance is simply convenience: each type of system has certain constraints (e.g. conservation of energy/mass for a closed system) that simplify the model.

The Helmholtz free energy is good for systems at constant V and P, the Gibbs free energy is preferred for systems at constant P and T.

Urmi Roy

Hmm.... the concept of mechanical energy when introduced into thermodynamics seems to be very ambiguous!

What I originally asked was how many state veriables completely define the thermodynamic state....entropy and enthalpy are determined by the other three,so I guess the answer to my original question,as my wan said, is 3.

Btw,why is it that "state space has an odd number of dimensions, whereas the mechanical state space has an even number" is important to note?

But if it isn't quasi-static,we can't even plot the resulting P-V diagram,as we can't define the Pressure for non-equilibrium states!

Density of mass. my wan gave a good answer to this question.

Okay,so would it be right to say that for most of our experiments on adiabatic processes,isobaric processes and isochoric processes are conducted with isolated systems wheras for isothermal process,we need a closed system?

Thanks for the help,Andy Resnick!

Andy Resnick

Not really- the work (mechanical energy) is just the PV term, oftentimes considered as P dV.

Not quite- entropy and enthalpy also require the heat (Q), they aren't completely specified by specifying P, V, and T.

As to the difference in even- versus odd- dimensioned phase space, I can only give a partial answer: in mechanical theories, phase states with an even number of dimensions have a symplectic structure: there are conjugate pairs of variables (i.e. position and momentum, q and p). Odd-dimensioned spaces have a contact structure:

http://en.wikipedia.org/wiki/Symplectic_vector_space
http://en.wikipedia.org/wiki/Contact_geometry


You are raising an interesting question, regarding the appropriateness of thermodynamic variables in the context of non-equilibrium conditions. Thermodynamics is a continuum theory- it doesn't really apply to dilute gases.

DrDu

Andi, what do you say! To specify the equilibria of the simplest systems two out of the three variables p, V, T are sufficient. The heat is not needed to specify U uniquely. Of course there are more complicated systems which need more variables for a complete specification of the equilibrium state, so there may also be electric and magnetic field variables, stress and strain etc. Thermodynamics does not fix this number but one has to find them for the specific system. Hence one sometimes says that thermodynamics is a meta-theory.

Andy Resnick

Please see, for example,

http://arxiv.org/abs/math-ph/0703061

For a single phase, the 5 variables are pressure/volume, temperature/entropy, and U.

alynfazlin

want to ask a question..is it true the absolute pressure in a liquid of c0nstant density d0ubles when the depth is d0ubled?explain please..