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![[Post New]](/templates/default/images/icon_minipost_new.gif) 16 Mar 2007 20:26:15 IST
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Two identical containers joined by small pipe initially contain the same gas at pressure 'p' and temperature T. One container is now maintained at the same temperature while the other is heated to 2T. The common pressure of the gases will be a. 3/2 p b. 4/3 p c. 5/3 p d. 2 p
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nupur.. |
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17 Mar 2007 12:10:56 IST
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plz reply yaar!!!1
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nupur.. |
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17 Mar 2007 12:18:34 IST
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2P
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ARUN
TRY,TRY,TRY TILL U SUCCEEEEEEEEEEEEEEEEEEED
YOU CAN DO WHATEVER U WANT,JUST TRY
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17 Mar 2007 13:31:03 IST
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Is the ans 3/2 p
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17 Mar 2007 18:10:30 IST
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nopes!!! the ans is B.
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nupur.. |
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17 Mar 2007 18:22:17 IST
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i think i may b able 2 ans it
let containers be A & B
the pressures in them will be p & 2p
ratio is 1:2
common press will be= (2xp)+(1x2p)/(1+2)=4p/3
it much like centre of mass or cord -geometry
but i am not sure about it so i wouldnt ask u to rate me.
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18 Mar 2007 15:24:38 IST
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Here is the solution
http://blogambuj.googlepages.com/sol.png
Click on above link to open solution.
sorry for my hand writing bad but readable.
anyways cheers
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IIIT Allahabad |
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21 Mar 2007 12:11:29 IST
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lets assume that first pressure is p...
and volume is v.... now use pv=nRT for both rooms... no of moles are same
Now as temp changes... but volume remains constant.. but no. of moles will change for each container but . their sum equal to 2n...
let now pressure be p1.. f
for container one p1v=n1RT
for container two p1v=n2R2T
now n=n1+n2..
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21 Mar 2007 13:13:44 IST
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well i mnt sure of this but i guess d ans is 2p this is bcos P is directly proportional to temp acc to amonton's law
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23 Mar 2007 01:56:56 IST
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Answer is b.
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