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This is kind of embarassing, but for some reason I can't do physics for shit... My teacher explains some things and does some easy problems, talking about them being "an insult to intelligence" and other people seem to get this but I don't. Maybe I just take bad notes :/
So, a few basic questions to help me on some of my homework:
For the equation PV=nRT, T is always Kelvin, right?
If you're calculating pressure on something underwater, do you add 101.3 atm to the pressure to get total pressure? Is there a scenario where you don't do that?
"An auditorium has dimensions 6 x 7 x 52 meters, universal gas constant (R) is 8.31451 J/K x mol, pressure is 1 atm, 20 degrees C, find moles of gas."
So I have V = 2148m^3, have T = 293, but what is P? Do I substitute 101.3 kPa for 1 atm? I tried both variations for P in the equation PV=nRT, but both were wrong... When do you substitute 101.3 kPa for atm?
Additionally, in general, does anyone have any tips on how to have a physics-geared mind? I get the problems wrong the first time almost always, and when I do get it right the knowledge doesn't seem to carry with me to the next problem.
Thanks in advance.
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Calgary25954 Posts
... First link on google.
1 atm = 1.01325 bar = 101.3 kPa = 14.696 psi (lbf/in2)= 760 mmHg =10.33 mH2O = 760 torr = 29.92 inHg = 1013 mbar = 1.0332 kgf/cm2 = 33.90 ftH2O
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the way i learned it... the universal gas constant (R) is
8.314472 L · kPa · K-1 · mol-1
or "kpal over molk"
so use those units: kPa, L, mol, K
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United States24496 Posts
On November 12 2006 22:59 Newbistic wrote:
This is kind of embarassing, but for some reason I can't do physics for shit... My teacher explains some things and does some easy problems, talking about them being "an insult to intelligence" and other people seem to get this but I don't. Maybe I just take bad notes :/
Don't feel bad. Most teachers haven't even learned yet that the way to teach someone something isn't to explain it to them.
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If you're working with volume at m^3 you use Pa not kPa, if you have the pressure at kPa you just have to multiply it for 10^3 to get Pa, If you use dm^3 then you use atm... it's kinda simple actually... ho and the contant of gases R = 8.314 is for when you use Pressure = Pa and Volume =m^3, if you're working with Pressure = atm and Volume = dm^3 then you must use R= 0.0821, this is cause R depends on the units you're working with (if you manipulate the expression R=PV/nT) i don't recall now how you get this second value for R (you could manipulate the expression knowing all the values and consider R the unknown value and get it from this last expression but there's just another way to prove it trust me  ) i just know it cuz i've had basic thermodynamics a year ago and this was something really trivial ... Anyway on to the next problem:
If you're on an environment that's already under pressure you just add the pressure you apply to the total pressure it already has, meaning, Ptotal= Penvironment + Papplied, now i don't know which pressure you're working underwater (Penvironment)... in fact that's hydrostatics it depends on the height of water above the object, i remember you have about 2 fundamental equations to get that but i don't remember :S... that was long time ago...
Well i hope i've been of help i know i must've written a lot of mistakes etc but i apologise myself about that...
Cheers :D
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thedeadhaji
39489 Posts
yea just check the units for what exactly kPa stands for, then you want to just start cancelling out the units until you are left with what you are looking for.
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1) Just use SI-units to get everything right. That is pascal (not kilopascal) for pressure. Kelvin for temperature, m^3 for volume and so on.
2) Preassure is 1 atm at water surface. Then it increases linearly with 1 atm every 10 meters. so 1.5 atm @ 5 meters, 2 atm @ 10 meters, 11 atm @ 100 meters, and so on. (1 atm = 101 300 Pa)
3) P is 101 300 Pa. if you used 101,3 instead you will get a factor 1=10^3 wrong.
Basicly you use Pa for most calculations, and atm when you want to get a feeling for how preassure it is. (wtf is 200 000 pascal?? Oh, 2 atm, omg, that's double normal preassure!!)
4) You have to understand the problem. If you are just learning the calculations, without understanding what the equations mean, then that is the problem. Otherwise, logic thinking is something you train. Think always one step at a time. Clean steps!
First understand the problem!
What do I know?
What do I want to know?
What tools do I have?
When can I use these tools?
How should I use these tools?
Take it easy, relax. Think.
good luck!
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Australia3818 Posts
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On November 12 2006 22:59 Newbistic wrote:
This is kind of embarassing, but for some reason I can't do physics for shit... My teacher explains some things and does some easy problems, talking about them being "an insult to intelligence" and other people seem to get this but I don't. Maybe I just take bad notes :/
Additionally, in general, does anyone have any tips on how to have a physics-geared mind? I get the problems wrong the first time almost always, and when I do get it right the knowledge doesn't seem to carry with me to the next problem.
Thanks in advance.
First you have to have a mentality that you can do them, or else you'll be hindering yourself when you meet a question and think "I'll probably get this wrong, as always". You might not have that mental barrier but if you do you need to throw it away.
For physics problems nearly all the time its about using known variables in an equation to find an unknown. When you read a question, you should write down everything you know. Also, write down the variable that the question asks of you. Then look at the variables and figure out which equation links all of them together, then put in the known values to find out the unknown. The rest of it is just math, but I'm not sure if you're good at that.
As Cascade said you should use SI units for everything. That way it all works out nicely.
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EPT
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