Change in Volume of Argon at various pressure
Change in Volume of Argon at various pressure
(OP)
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!
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!





RE: Change in Volume of Argon at various pressure
RE: Change in Volume of Argon at various pressure
Good luck,
Latexman
Need help writing a question or understanding a reply? forum1529: Translation Assistance for Engineers
RE: Change in Volume of Argon at various pressure
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.
RE: Change in Volume of Argon at various pressure
Good luck,
Latexman
Need help writing a question or understanding a reply? forum1529: Translation Assistance for Engineers
RE: Change in Volume of Argon at various pressure
RE: Change in Volume of Argon at various pressure
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!
RE: Change in Volume of Argon at various pressure
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
Need help writing a question or understanding a reply? forum1529: Translation Assistance for Engineers
RE: Change in Volume of Argon at various pressure
Good luck,
Latexman
Need help writing a question or understanding a reply? forum1529: Translation Assistance for Engineers
RE: Change in Volume of Argon at various pressure
Good luck,
Latexman
Need help writing a question or understanding a reply? forum1529: Translation Assistance for Engineers
RE: Change in Volume of Argon at various pressure
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!
RE: Change in Volume of Argon at various pressure
RE: Change in Volume of Argon at various pressure
RE: Change in Volume of Argon at various pressure
\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
RE: Change in Volume of Argon at various pressure
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: Translation Assistance for Engineers
RE: Change in Volume of Argon at various pressure
RE: Change in Volume of Argon at various pressure
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.
RE: Change in Volume of Argon at various pressure
395.9 K, now resulting in a volumen of 412.1 L, still a small deviation from NIST figures.
RE: Change in Volume of Argon at various pressure
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.