Heating a gas, then expanding.

[Fool]

Heating a gas, then expanding.

This thread is to separate out that question from the, "Why a temperature differential.", thread.

The following appears to be Jack's original question.

To check my understanding of this. If a fluid is heated it will keep expanding until it reaches ambient temperatures again?
So assume you have two separate liters of fluid in a closed fully insulated space and you heat one up, you expand it until it reaches starting temperature again. Now you have one with a bigger volume than the other, the same temperature and pressure but lower density?

I cross posted with him and would like to apologize, as it looks a little like I was responding to he, when in fact I was responding to another poster. I think Jack recognizes that. I won't rehash that here, as it's off topic. The apology and explanation was appropriate, and missing, so I'll put it here.

Jack I have no problems with you, please keep the questions coming. That response was for another poster. I recognize that you are here to help and understand. Maybe to teach others back home, too.

Okay, then let's continue. Jack's first question was actually two that are related :

"To check my understanding of this. If a fluid is heated it will keep expanding until it reaches ambient temperatures again?"

And:

"So assume you have two separate liters of fluid in a closed fully insulated space and you heat one up, you expand it until it reaches starting temperature again. Now you have one with a bigger volume than the other, the same temperature and pressure but lower density?"

The first one is posted as a statement, with a question mark at the end appearing to indicate that he wants to learn more if need be.

I would like to comment on that statement. Generally when a gas is expanded after heating has stopped, its temperature reduces from a process of adiabatic work. However, a few details need to be pointed out.

Expansion only happen if the system let's it. Merely heating a gas won't always lead to expansion. Case in point for any sealed tank, such as an oxygen or propane bottle, heating it will only raise the temperature and pressure, not the volume. The bottle constrains the gas to a constant volume. Constant volume means zero expression, or compression.

Of course the gas can have other expansion processes than adiabatic with work, since you've clarified, and specified an adiabatic with work expansion process, I will only explain that point.

You appear well aware that heating and adiabatic are opposites. So I won't bemoan that.

The second question seems to be setting up an experiment with one vessel space used as a 'control' and remaining at Tc ambient. And another that you have heated to Th. Then you have forced it to expand adiabatically with work until the gas is again back down to Tc. You now are asking what state it is in?

You make the correct point that it will have a larger volume and lower density. I agree. Much much lower density and larger volume. And lower pressure. Temperature back to Tc.

I supplied a couple of different ways to calculate that using gas laws. If you'd like, I would repost them here and explain them more simply. Mathematics is only learned if the explanation matches the person.

You have adequately asked a question and clarified it many ways. I hope this thread answers it more clearly, and allows more questions and pondering. Cary on men and women.

[VincentG]

It would help to have some index of proper terms.

To me, for the context of this forum, a "free expansion" is air contained within a piston and cylinder, pushing outwards against a buffer pressure of 1ATM. This would simulate the middle ground between heat pump and engine, the bare minimum "work" the gas must do to expand, without any extra "work" being taken from or done to the gas.

To say "free expansion" into a vacuum, seems not only inapplicable but also misleading as expansion into a vacuum does not an engine make.

[Jack]

Thanks for condensing it all into one post. This all makes it very clear.

From the other thread I learned an interesting new tidbit that I still have to fully "implement" into my own understanding.
From what I gathered before, temperature = internal energy = molecular kinetic energy. And the adiabatic expanding in my question doesn't actually lower the internal energy, but merely gives the molecules more space to move around in.

If this is true then temperature is a ratio between internal energy and volume?

This might be very basic knowledge for a lot of you guys, but it's quite eye opening for me hehe.

[Fool]

VincenG, a free to move piston, assuming negligible mass, such as a balloon, when heated will expand as the heat transfers in. It will expand keeping Pin = Pout. The temperature and volume will be higher, and the pressure will be constant.

In Matt's Calculator, isobaric can be selected. I haven't played with that setting.

Jack, if temperature stays constant, internal energy stays constant.

∆U=MCv∆T

M mass, Cv heat coefficient, MCv are the slope of a line, a constant.

[Jack]

But if you expand a gas, the temperature goes down, right?
Does that mean the internal energy goes down?

Intuitively right now this feels like internal energy is dependent on density then. Before I said volume, but that doesn't seem right.

[Jack]

I'm messing this up again haha. Temperature is a ratio of internal energy and density?

[VincentG]

Jack you will find that you will never get a straight answer to this for your practical brain. The math is a reality but that doesn't mean reality is the math.

It can be fun to banter about(until some people get out of hand) but it really doesn't change how an engine works either way.

Build cool stuff and have fun.

[Fool]

Jack, it depends on how you look at it. The equations can be altered using algebra to get the function expressed in the variables you want.

Density D = Mass÷Volume=M/V

Solve it for mass

M=DV

Internal energy U=MCvT

Substitute For M:

U=DVCvT

Internal energy as a function of Density, Volume, and Temperature. Cv being a constant for a the gas used. The value DV is also constant, for the quantity of gas sealed in, because M is constant. Essentially saying internal energy is a function of the Temperature only.

Further investigation:
PV=MRT

Substitute for M here too:
PV=(DV)RT

Solving it for T:

T=P/(DR)

Putting it into:
U=(D/V)CvT

U=(D/V)Cv(P/(DR))

Simplifying:

U=(CvP)/(RV)

Density cancels and pressure is added. It's now a function of pressure over volume, P/V.

Pressure increases internal energy increases.

Volume increases internal energy decreases, and density decreases, total mass stays the same. Pressure decreases.


I hope this helps. Please ask for a more detailed explanation.

VincentG, building something cool, yes. Good plan.

[VincentG]

Density cancels and pressure is added. It's now a function of pressure over volume, P/V.

Pressure increases internal energy increases.

Volume increases internal energy decreases, and density decreases, total mass stays the same. Pressure decreases.

Its amusing that when I was making this case I was wrong.

[Jack]

I'll study it a bit more carefully when I have a bit more time for it. But thanks for taking the time to explain.

Hopefully we can all keep building cool stuff!

[Fool]

VincentG, I was wrong too. I found out when I did the math and changed the post before posting. It seems logical until the mathematics clobbers you.

[matt brown]

VincentG, I was wrong too. I found out when I did the math and changed the post before posting. It seems logical until the mathematics clobbers you.

Hence my common bunny fart about thermo..."some is intuitive and some is counter-intuitive"

Thermo is the most confusing subject I know of where conventional logic offers little solace. No doubt, extensive math knowledge helps for calculations, but a "logical" math mindset often hinders grasping many "illogical" thermo concepts. I grew up in a tech rich environment and studied thermo from a very young age, so I can usually walk thru the quick sand. However, once past your formative years (college age) thermo appears esoteric and akin teaching an old dog new tricks.

A - here's another tidbit I've learned over years of thermo discussions:
B - what you learn often depends on how you came to learn it
C - after grasping B, reread A

[Fool]

Science is physics, physics is math, so science is math. It is possible for anyone to learn math/science/physics if they work diligently enough. We are all bozos on this bus.

[Tom Booth]

Science is physics, physics is math, so science is math. It is possible for anyone to learn math/science/physics if they work diligently enough. We are all bozos on this bus.

Science is not math, or does not necessarily include or require any mathematics.

[Fool]

Good luck with that. I would like to see a relevant physics sentence that doesn't include or imply a mathematical concept.