What's new
Fantasy Football - Footballguys Forums

Welcome to Our Forums. Once you've registered and logged in, you're primed to talk football, among other topics, with the sharpest and most experienced fantasy players on the internet.

Gas pressure question (1 Viewer)

chet

chet

Footballguy
Suppose you have two steel tanks with pressure gauges, each with a volume of 100 gallons.

They are connected with a pipe that has a cut off valve.

Each tank has air in it.

One tank has indicated 100 PSI.

The second tank has indicated 0 PSI.

If you open the valve, what will the pressures be after equilibration?

Is pressure linearly related to amount of air in a set volume?

 
The caveat is your gauge is measuring relative pressure, which probably means there is actually 14.7 psi and not actually zero. 

 
chet said:
Unpack this please.
Click to expand...
It becomes a bit of a different problem if the other tank is a vacuum verses having one atmosphere of pressure (which is what a relative guage calls zero).   In the case of a vacuum you are doubling the volume that the air in the second tank would later fill, so the answer would be half.   For the relative gauge where your real pressure is 14.7 psi in one tank and thus 114.7 psi in the second tank, the problem becomes more complex and you would have to use the equation in the first post to calculate.   I am in a car, so I can't.  

 
chet said:
Suppose you have two steel tanks with pressure gauges, each with a volume of 100 gallons.

They are connected with a pipe that has a cut off valve.

Each tank has air in it.

One tank has indicated 100 PSI.

The second tank has indicated 0 PSI.

If you open the valve, what will the pressures be after equilibration?

Is pressure linearly related to amount of air in a set volume?
Click to expand...
50 psi.  Yes.

jon_mx said:
It becomes a bit of a different problem if the other tank is a vacuum verses having one atmosphere of pressure (which is what a relative guage calls zero).   In the case of a vacuum you are doubling the volume that the air in the second tank would later fill, so the answer would be half.   For the relative gauge where your real pressure is 14.7 psi in one tank and thus 114.7 psi in the second tank, the problem becomes more complex and you would have to use the equation in the first post to calculate.   I am in a car, so I can't.  
Click to expand...
You're over thinking it.  Both gauges are likely to be calibrated for zero at atmosphere.

 
tonydead said:
50 psi.  Yes.

You're over thinking it.  Both gauges are likely to be calibrated for zero at atmosphere.
Click to expand...
No, I don't think I am.  For the same reason you must use a Temperature scale where absolute zero is zero, you must use a pressure scale where zero means zero.  I need to put this on paper to convince myself.  The gauge would read something close to 50.

 
jon_mx said:
chet said:
Unpack this please.
Click to expand...
It becomes a bit of a different problem if the other tank is a vacuum verses having one atmosphere of pressure (which is what a relative guage calls zero).   In the case of a vacuum you are doubling the volume that the air in the second tank would later fill, so the answer would be half.   For the relative gauge where your real pressure is 14.7 psi in one tank and thus 114.7 psi in the second tank, the problem becomes more complex and you would have to use the equation in the first post to calculate.   I am in a car, so I can't.  
Click to expand...
Said another way, there's relative and absolute pressures.

Absolute, 0 means a vacuum.

Relative, 0 means atmospheric pressure which is a higher pressure than a vacuum.  At sea level, atmospheric pressure is about 14.7 PSI above vacuum.  So 0 PSI relative = 14.7 PSI absolute.    

 
jon_mx said:
No, I don't think I am.  For the same reason you must use a Temperature scale where absolute zero is zero, you must use a pressure scale where zero means zero.  I need to put this on paper to convince myself.  The gauge would read something close to 50.
Click to expand...
That's because temperature is on the temperature side of the equation and Kelvin represents the total kinetic energy of the particles in a mole.  He doesn't even have to use that if the temperatures arehe same, Celsius or Kelvin.  P1V1 = P2V2  It doesn't matter as long as the pressure gauges are calibrated the same which is a fair assumption.

 
tonydead said:
That's because temperature is on the temperature side of the equation and Kelvin represents the total kinetic energy of the particles in a mole.  He doesn't even have to use that if the temperatures arehe same, Celsius or Kelvin.  P1V1 = P2V2  It doesn't matter as long as the pressure gauges are calibrated the same which is a fair assumption.
Click to expand...
PV = nRT  

n is the number of gas molecules

If the readings are absolute PSI, then the 0 PSI tank has a vacuum and n=0.  When you combine tanks you have nRT from the full tank and add 0 from the other tank because of that 0.

What if you have relative pressures?   Then 0 means you have 1 atmosphere of gas. That means there are gas molecules in the "empty" tank and n is non-zero.  So when you add the tanks together, you are not just doubling the volume, you are also adding additional gas molecules.  You cannot just neglect the nRT side of the equation now because n from the additional tank is non-zero.

Make sense?

 
Last edited by a moderator:
I calculate 395,672 Pa or 57.4 psi.

ETA - this was wrong.  I made an error as noted further down the thread.  Correct answer is 50 psi.

 
Last edited by a moderator:
To elaborate...as others have stated, the pressure readings given are gauge pressure which is the difference between absolute pressure (zero for a perfect vacuum) and atmospheric pressure.  Both tanks are filled with air with equal volume but different pressures.  That means there is more air in one tank than the other.  Use PV = nRT to figure out how much air is in each tank (n).  To use that equation, you should have absolute pressure.  Once you have the total n, you can then calculate the pressure for that amount of air spread out in 2*Volume. 

 
Last edited by a moderator:
GregR said:
PV = nRT  

n is the number of gas molecules

If the readings are absolute PSI, then the 0 PSI tank has a vacuum and n=0.  When you combine tanks you have nRT from the full tank and add 0 from the other tank because of that 0.

What if you have relative pressures?   Then 0 means you have 1 atmosphere of gas. That means there are gas molecules in the "empty" tank and n is non-zero.  So when you add the tanks together, you are not just doubling the volume, you are also adding additional gas molecules.  You cannot just neglect the nRT side of the equation now because n from the additional tank is non-zero.

Make sense?
Click to expand...
The other tank had 1 atmosphere of gas too, before it was charged and now is reading 100 psi.  

 
Galileo said:
To elaborate...as others have stated, the pressure readings given are gauge pressure which is the difference between absolute pressure (zero for a perfect vacuum) and atmospheric pressure.  Both tanks are filled with air with equal volume but different pressures.  That means there is more air in one tank than the other.  Use PV = nRT to figure out how much air is in each tank (n).  To use that equation, you should have absolute pressure.  Once you have the total n, you can then calculate the pressure for that amount of air spread out in 2*Volume. 
Click to expand...
The tanks have to come to equilibrium.  Two tanks reading 57.4 psi?  You have more air molecules than you started with.     Scratch that.  I though the gauges were calibrated at zerorwith 1atm?  Your answer is real pressure not what the gauges are reading.

 
Last edited by a moderator:
tonydead said:
The tanks have to come to equilibrium.  Two tanks reading 57.4 psi?  You have more air molecules than you started with.
Click to expand...
No, I don't

n = PV/RT

Assuming STP...

P1 = 0 psi (gauge) = 101,325 Pa

P2 = 100 psi (gauge) = 790,829 Pa

V = 100 gal = 0.003785 m^3

R = 8.314 J/mol K

T = 273 K

# of mol of air in the 0 psi (101,325 Pa) tank is  101325(0.003785 ) / 8.314 8 (273) = 0.17 mol   

For the tank at 100 psi there is 790829(.0037785) / 8.314(273) =  1.32 mol.

So there is a total of 1.49 mol of air in the combined tanks.  That value of n is spread over a volume of 2V c

P = nRT/2V = 1.49*8.314*273 / 2*0.003785 =  446,748 Pa or 64.8 psi    Absolute pressure which is indeed 50 psi gauge pressure.

ETA - the above is now correct, although it does represent a long winded approach with details spelled out that become unnecessary in the given scenario.   

 
Last edited by a moderator:
Galileo said:
No, I don't

Assuming STP, the # of mol of air in the 0 psi (101,325 Pa) tank is  101325(0.003785 ) / 8.314 8 (273) = 0.17 mol    For the tank at 100 psi there is 689476(.0037785) / 8.314(273) = 1.15 mol.

So there is a total of 1.32 mol of air combined in the combined tanks.  That value of n spread over a volume of 2V corresponds to a pressure of 395,777 Pa or 57.4 psi 
Click to expand...
Hang on there, I revised my post.  How can you assume an initial atmosphere of pressure in a tank that reads 0 and at at the same time take a tank that reads 100psi at face value (see the bold)?  A logical assumption is that the guages are calibrated the same therefore you need to add another 14.7 psi to your 100 psi tank.  Then when you are done subtract an atmosphere off of your calculation pressures to answer what his question was: What will the pressure be (I interpret as what would the gauge read)?

 
Last edited by a moderator:
Big assumption here that the gauges are calibrated to zero at 1 atmosphere of pressure. What if you live in Denver?

 
tonydead said:
Hang on there, I revised my post.  How can you assume an initial atmosphere of pressure in a tank that reads 0 and at at the same time take a tank that reads 100psi at face value (see the bold)?  A logical assumption is that the guages are calibrated the same therefore you need to add another 14.7 psi to your 100 psi tank.  Then when you are done subtract an atmosphere off of readings (which should be the same) to answer what his question was: What will the pressure be (I interpret as what would the gauge read)?
Click to expand...
You are correct that I reported absolute pressure above with the 57.4 psi value.   ZERO gauge pressure is 1 atm = 101,325 Pa.  You correctly point out my error for the second tank.  The gauge pressure is 100 psi gauge pressure so the absolute should be 114.7 psi = 790,829 Pa... I need to re-calculate above.  Ooops...

 
P1 = 114.7 psi

V1 = 100 gallons

P2 = 14.7 psi

V2 = 100 gallons

n3 = 8.80n2

P3 = Answer to question

V3 = 200 gallons

200P3/8.80 = 100P2 = 1470

P3 = 64.7 psi

Gauge reads 50 psi.

A bit surprised it worked out that way.  

 
Last edited by a moderator:
Galileo said:
No, I don't

n = PV/RT

Assuming STP...

P1 = 0 psi (gauge) = 101,325 Pa

P2 = 100 psi (gauge) = 689,476 Pa

V = 100 gal = 0.003785 m^3

R = 8.314 J/mol K

T = 273 K

# of mol of air in the 0 psi (101,325 Pa) tank is  101325(0.003785 ) / 8.314 8 (273) = 0.17 mol   

For the tank at 100 psi there is 689476(.0037785) / 8.314(273) = 1.15 mol.

So there is a total of 1.32 mol of air in the combined tanks.  That value of n is spread over a volume of 2V c

P = nRT/2V = 1.32*8.314*273 / 2*0.003785 =  395,777 Pa or 57.4 psi    (slight difference from my earlier answer above from rounding differently...not significant.

edited to clean up my work a little...
Click to expand...
R's and T's were constants and canceled out.  

 
jon_mx said:
P1 = 114.7 psi

V1 = 100 gallons

P2 = 14.7 psi

V2 = 100 gallons

n1 = 8.80n2

P3 = Answer to question

V3 = 200 gallons

200P3/8.80 = 100P2 = 1470

P3 = 64.7 psi

Gauge reads 50 psi.

A bit surprised it worked out that way.  
Click to expand...
I just fixed my math above as well... I am a bit surprised to.  

 
:thumbup:   My brain experiment looked like this:

atm plus zero-------/-----atm plus 100psi

atm plus 50psi--------------atm plus 50psi

There is no energy introduced into the closed system.

 
Ketamine Dreams said:
P1V1 = P2V2
Click to expand...
But you are combining two different samples beginning in different states to form a third.  You can tap this general idea, but not without accounting for the difference in amount of matter in the two samples as jon_mx's solution does above.  

 
Galileo said:
I calculate 395,672 Pa or 57.4 psi.

ETA - this was wrong.  I made an error as noted further down the thread.  Correct answer is 50 psi.
Click to expand...
 Let me explain it another way. Let's assume I run a business that fills bicycle tires with air. The more psi I have the more tires I can fill, right? And also tanks are free and limitless (they are just containers for our closed system in this example).  If adding tanks increased psi why wouldnt I just do that over and over? 

 
Last edited by a moderator:
What is the length of the pipe. Diameter of the pipe? Where is the valve? Halfway? 

If we are going to nerd this up, let's really nerd it up. 

 
Galileo said:
Ketamine Dreams said:
P1V1 = P2V2
Click to expand...
But you are combining two different samples beginning in different states to form a third.  You can tap this general idea, but not without accounting for the difference in amount of matter in the two samples as jon_mx's solution does above.  
Click to expand...
I think that since the two individual tanks are the same volume and temp, you could simplify it by assuming all the initial gas and it's corresponding pressure was in one of the two tanks.

V1 = 100 gal (tank 1)

P1 = 100 psi + 14.7 psi (1 atm from tank 1) + 14.7 psi (actually from tank 2 but assumed to be in tank 1) = 129.4

V2 = 200 gal (both tanks together)

P2 = ?

P1V1 = P2V2

(129.4)(100) = (P2)(200)

P2 = 64.7 psi   -----   pressure gauge reads 64.7-14.7 = 50

I think this also makes it clearer why the initial hidden atmospheric pressures shouldn't change anything in the final result.

If PG = pressure gauge reading  and  A= atmospheric pressure

then P1 = PG readings of the two tanks plus atmospheric pressure for each of the tanks assumed all in one tank

and  P2 = the final PG reading plus the atmospheric pressure for the combined tank

P1V1 = P2V2

(PG1 + 0 + A + A)(100) = (PG2 + A)(200)  

(PG1 + 2A) = (PG2 + A)(2)

PG1 + 2A = 2(PG2) +2A

PG1 = 2(PG2)

 
Everybody is wrong so far, PV=ZnRT where z is the compressability factor of air.

z will be smaller the greater the pressure, so the tank with a higher pressure has a lower z value to start with.

It will be close to 50, but not exact.

Edit:

Wikipedia. Good enough for this discussion.

https://en.wikipedia.org/wiki/Compressibility_factor

 
Last edited by a moderator:
MTskibum said:
Everybody is wrong so far, PV=ZnRT where z is the compressability factor of air.

z will be smaller the greater the pressure, so the tank with a higher pressure has a lower z value to start with.

It will be close to 50, but not exact.

Edit:

Wikipedia. Good enough for this discussion.

https://en.wikipedia.org/wiki/Compressibility_factor
Click to expand...
Making a common assumption to get to an answer which gets you to within one percent of the correct answer is not wrong.   Given the information provided, it was the best solution.   

 
You must log in or register to reply here.

Similar threads

Anarchy99
Debating About Reporting A Service Company
2
Replies
84
Views
3K
Mitchie
M
S
Am I (we) the problem (AITA) - advice from the collective wisdom of the FFA
2 3
Replies
129
Views
3K
GroveDiesel
GroveDiesel
GordonGekko
Energy Crisis: Has Your Electricity/Gas Bill Gone Completely Out Of Control Recently? (16/30)
Replies
6
Views
441
STEADYMOBBIN 22
STEADYMOBBIN 22

Users who are viewing this thread

Total: 2 (members: 0, guests: 2)
Top