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Thermal Physics

Blazing goIITian

Joined:	25 Dec 2007
Post:	1104

28 Dec 2007 21:35:19 IST

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559

An iit level Q. plss help!!

None

A cylindrical tube with adiabatic walls and fitted with a diathermic
separator. Te separator can be slid in the tube by external mechanism.An ideal
gas is injected in the two sides at equal pressure and equal temperature.The
separator remains in the equilibrium at the middle.It is now slid to a position
where it divides the tube in the ratio 1:3.Find the ratio of pressures in the
two part of the vessel.

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Comments (3)

abhishek sinha

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Joined: 18 Dec 2007

Posts: 926

28 Dec 2007 21:46:54 IST

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apply pv power gamma = const

so , p1 : p2 = 3^ ( gamma ) =6.26

gamma = 1.67 for monoatomic gas

#$nishi$#

Cool goIITian

Joined: 26 Dec 2007

Posts: 97

1 Jan 2008 20:31:43 IST

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according to question tube is divided in ratio 1:3,

ratio of volume =1:3

from,

PV=nRT

therefore,P2\P1=V2\V1=3:1

abhishek sinha

Forum Expert

Joined: 18 Dec 2007

Posts: 926

1 Jan 2008 20:34:17 IST

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See, the parts are seperated by diatharmic walls ( no heat exchange )

So their final temp would not be equal .

Hence the method used by u is wrong !!!!!!!!!!!!!

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