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Cylinder gas, pressure question

  • Thread starter yopy
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  • #1
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http://img15.imageshack.us/img15/9746/21478397.png [Broken]

I know how to setup the entire problem i just dont know how to calculate the new pressure after the piston is compressed.
 
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Answers and Replies

  • #2
LowlyPion
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I know how to setup the entire problem i just dont know how to calculate the new pressure after the piston is compressed.
PV = nRT

n, R, T remain the same. They give you everything else.
 
  • #3
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PV = nRT

n, R, T remain the same. They give you everything else.
how do i calculate the number of moles?
 
  • #4
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maybe p1v1/t1 = p2v2/t2 is more appropriate?
 
  • #5
43
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maybe p1v1/t1 = p2v2/t2 is more appropriate?
i dont think so, this is actually from my physics class and we never covered that
 
  • #6
76
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ok fair enough
just out of interest is the answer 234375 Pa?
 
  • #7
mgb_phys
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What Rory is saying is that PV = nRT is the general equation but nRT and are constant
So p1v1 = p2v2, rearrange a little and you have the answer
 
  • #8
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ok fair enough
just out of interest is the answer 234375 Pa?
not sure, has yet to be handed in
 
  • #9
76
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ahhh yes even more simple! forgot the temperature remained constant!
ok fair enough, would you be as kind enough to tell us the answer as i am revising for my physics A-level at the minute so all questions are useful!!!
 
  • #10
mgb_phys
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just out of interest is the answer 234375 Pa?
No.

The volume goes from 0.750 -> 0.480 m^3, the pressure starts at 1.50 x10^5 Pa
so the final pressure is (0.750/0.480) * 1.50 x10^5 Pa = 2.34 x10^5 pa

You understand why your answer is wrong?
 
  • #11
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No.

The volume goes from 0.750 -> 0.480 m^3, the pressure starts at 1.50 x10^5 Pa
so the final pressure is (0.750/0.480) * 1.50 x10^5 Pa = 2.34 x10^5 pa

You understand why your answer is wrong?
lol thats the answer i got, except yours is in standard form! lol
 
  • #12
mgb_phys
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This is a 'physics' A level not maths
2.34 x10^5 is correct 234375.000000000000000000000 pa is wrong - whatever your calculator says!

Hint, whats the precision of the initial pressure measurment
 
  • #13
76
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oooohhh slightly harsh!!
but never the less thats fair enough!
no wonder my physics teacher is always annoyed with me for leaving whole answers!!!
 
  • #14
43
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maybe p1v1/t1 = p2v2/t2 is more appropriate?
yea this is the correct way to do it, thanks
 
  • #15
43
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ahhh yes even more simple! forgot the temperature remained constant!
ok fair enough, would you be as kind enough to tell us the answer as i am revising for my physics A-level at the minute so all questions are useful!!!
http://wug.physics.uiuc.edu/cc/IAState/Phys221/spring/homework/written%20homework/index.html [Broken]


have at it.
 
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