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#1 lucy skies

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Posted 11 September 2008 - 10:20 PM

Hey, everyone. I'm a freshman Ch E major and I need some help with Thermodynamics.

Question: A 1-m^3 rigid tank has propane at 100 kPa, 300K and connected by a valve to another tank of 0.5m^3 with propane at 250kPa, 400K. The valve is opened and the two tanks come to a uniform state at 325K. What is the final pressure?

Explanations would be very helpful.

#2 siretb

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Posted 12 September 2008 - 01:45 AM

Easy. You should have found in a book by yourself
we have low pressures, low temperatures ; ideal gas law will do

1st tank P1 V1 = n1 R T1 P pressure, T temperature , n number of moles, V1 volume, R a constant
that gives n1
2nd tank P2 V2 = n2 R T2
that gives n2

and then you consider you have one big tank, with the total
I let you do the calculation.

#3 lucy skies

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Posted 12 September 2008 - 10:47 PM

QUOTE (siretb @ Sep 12 2008, 01:45 AM) <{POST_SNAPBACK}>
Easy. You should have found in a book by yourself
we have low pressures, low temperatures ; ideal gas law will do

1st tank P1 V1 = n1 R T1 P pressure, T temperature , n number of moles, V1 volume, R a constant
that gives n1
2nd tank P2 V2 = n2 R T2
that gives n2

and then you consider you have one big tank, with the total
I let you do the calculation.


Which volume would you use? Do you add the volumes together? If you only use one volume for the overall big tank, why would you use that one? I plugged my numbers in as

P (V1+V2)=(n1+n2)®(T) = and got a ridiculous 93 Pa. No way that is the right answer.

#4 chenblue

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Posted 12 September 2008 - 11:56 PM

Hi, Lucy,
Your equation is correct, but I guess you mistook something during your calculation. I got a 139.9 kpa from your equation,
P(V1+V2)=(n1+n2)RT.

#5 lucy skies

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Posted 14 September 2008 - 10:12 AM

QUOTE (chenblue @ Sep 12 2008, 11:56 PM) <{POST_SNAPBACK}>
Hi, Lucy,
Your equation is correct, but I guess you mistook something during your calculation. I got a 139.9 kpa from your equation,
P(V1+V2)=(n1+n2)RT.

Thank you! I honestly appreciate the help.




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