Change in Volume of Argon at various pressure

[strongboes]

Hello,

I am struggling with a problem and will be so grateful if somebody can help after weeks of trying!

This is the scenario, please consider a reversible Adiabatic process, If I have an insulated balloon filled with 1kg or 25.03 moles of Argon gas then at 100kpa I make the volume to be 624.3 litres

Lets assume this balloon is in water, what I want to know is if I push this balloon down and increase the depth what volume and temp will the Argon inside balloon get to.

If I go to 200kpa or approx 10m deep, I know if temp stays the same that the volume will be 333.17 litres but the temp wont stay the same as it has been compressed and it is insulated, how do I figure out the temp and volume in this situation?

Please help!

 

[strongboes]

Sorry just to say that this question is out of my field and it is for a personal project/prototype I am creating. Thanks

 

[Latexman]

I'm glad you cleared that up, because it sounds like a thermo homework problem!

Good luck,
Latexman

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[strongboes]

Yes looking at it your right, it does sound homeworky, but it's not, it is for a potential energy storage solution and I cant say much more than that really.

My only option is to actually do this and record the temp and pressure as I can't figure the maths out, it is beyond me and I dont mind admitting that.

 

[Latexman]

There is a pretty good search engine at the site. It is under the title of the thread and between Forum and FAQs. I suggest a search on adiabatic gas compression or ideal gas compression. You should get many threads to look through for the equation.

Good luck,
Latexman

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[25362]

May be this link is of help:

http://webbook.nist.gov/cgi/cbook.cgi?ID=C7440371&Mask=4

​

 

[strongboes]

Latexman,

I have used a the adiabatic equation but I can only work out what happens if I change the volume from 1 value to another with a starting temp. I get out final temp and final pressure.

I cant work out how to do it so I have starting pressure and final pressure, what the vol and temp will be? I have searched many sites for literally weeks now. Are you able to help? Is there an equation for what I want to find out?

I know it is also called an Isentropic process and I think the Enthalpy stays the same which allows you to work out what I need, but when I look at the equation I really dont understand what I need to do. I need babying and I am willing to paypal someone a small sum if they can help!

 

[Latexman]

You know P1, T1, and V1. Right?

You know P2. Right?

You need the adiabatic index, k = Cp/Cv, for argon. I believe it's 1.4, but, please, look it up.

T2 = T1*(P2/P1)^((k-1)/k)

Now, argon is modelled pretty closely by the ideal gas law, so you can combine Boyle's and Charles's Law to get:

P1V1/T1 = P2V2/T2

and

V2 = V1*(P1/P2)*(T2/T1)

Good luck,
Latexman

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[Latexman]

Oh yeah, you have to use absolute pressures and absolute temperatures!

Good luck,
Latexman

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[Latexman]

There are other ways to solve it too. Have you seen this - Adiabatic Compression

Good luck,
Latexman

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[strongboes]

Ok Thankyou Sir.

I will now try and put that together. I have found the k value to be 1.67

so if I use kelvin value of 300, and pressure value of 100kpa assuming the balloon is just submerged, is that correct?

And a starting vol of 0.6243m3

P2 would be 200kpa for a depth of 10m

T2 = T1*(P2/P1)^((k-1)/k), 300*(200/100) ^I dont know what this symbol means? ((1.67-1)/1.67). so its 600 something 0.401


V2 = V1*(P1/P2)*(T2/T1), 0.6243*2*(T2/300)


Oh dear is that correct? obviously I cant do volume as I dont know T2 as I dont know what the symbol means? I have tried looking it up!

 

[strongboes]

Yes I have read that page 10 times but the equations I dont understand

 

[zekeman]

But why are you assuming the process is reversible?

 

[sailoday28]

Try the NIST website thermo properties of ARGON
\

http://www.nist.gov/data/PDFfiles/jpcrd363.pdf

Starting around pg 690 Pick your starting pressure and Temp to get density and entropy.
Then for isentropic process scan pages for the the final pressure. You then have final condition properties.

Regards

 

[Latexman]

zekeman - don't confuse him with the facts (of the real world)! He's struggling with the ideal world.

strongboes - are you an engineer? Even an engineering student knows what the math operator ^ means. It means "raised to the power of". Some symbologies use **, but that's old school. More explicitly, (600)^0.401. And, I am blindly using your numbers for this example.

Good luck,
Latexman

Need help writing a question or understanding a reply? forum1529

 

[strongboes]

Yes ok I know I look really silly. But doesnt that give me an answer of 13. And so with a starting temp of 300 i assumed i must have done it wrong?

 

[25362]

Using the link I gave you, and as sailoday28 explained, the isoentropic compression of 1 kg Ar gives the following results:

P₁ = 1 bara
T₁ = 300 K
V₁ = 624.3 L
S (constant entropy) = 0.9271 kcal kg^-1 K^-1
P₂ = 2 bara
T₂ = 395.9 K
V₂ = 412 L

Using the equations that Latexman gave you:

T₂ = T₁(P₂/P₁)^(k-1)/k = 300 (2/1)^0.401 = 396.1 K

V₂ = V₁(1/2)(395.9/300) = 411,9 L

Results almost equal to those tabulated by NIST.

 

[25362]

In the 2nd equation I should heave written 396.1 K instead of
395.9 K, now resulting in a volumen of 412.1 L, still a small deviation from NIST figures.

 

[racookpe1978]

Wait a minute: You either have an insulated balloon => In that case, the temperature of the gas inside the balloon and the balloon walls won't change with time (or change very slowly w/r to time compared to change in pressure outside the balloon or the movement of the gasses due to other things going on); or you don't have an insulated balloon.

If the balloon is not insulated, then the balloon walls almost immediately become the temperature of the water around the balloon (mass water = big, thermal coef of water = very big; mass balloon walls = very small, contact area of balloon wall/mass of balloon walls = very, very big.) Very shortly after the balloon walls are inserted into the water, the Ar will begin cooling down and begin to approach the water temperature as well.