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![[Post New]](/templates/default/images/icon_minipost_new.gif) 10 Apr 2008 11:15:52 IST
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A glass sphere of volume 7litres contains air at 300K. It is connected to a pipe filled with mercury as shown in the figure. At the start, the mercury meniscus is level with the bottom of the sphere on both arms of the pipe. The air in the sphere is now heated and the mercury level on the right arm rises by 5mm. If the cross sectional area of the pipe is 10cm^2, what is the new temperature of the air ?
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10 Apr 2008 17:01:36 IST
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=> Mercury level goes down by 5 mm below the bottom of the sphere.
Now because the right end is open to atmosphere so pressure initially in the sphere is also equal to the pressure at the top most point of the right end = 76mm of HG
Increase in pressure = 5mm of mercury.
No. of moles is constant.So equate them with new values of P and V
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Let us learn to dream, gentlemen, and then perhaps we shall learn the truth.
- August Kekule |
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10 Apr 2008 18:59:30 IST
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Volume of air is constant here.
Hence, using ideal gas equation, we get VdP = nRdT
Now, find dP, as you have been given the increase in the head. Hence, find dT
Required answer is 300+dT
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Will nip in at times to solve problems :)
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10 Apr 2008 19:16:58 IST
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Impossible To be Impossible is Impossible |
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10 Apr 2008 19:23:05 IST
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Continuing my solution, since people are posting different approaches :)
(76 * 7)/R 300 = 76.5 * (7.005)/RT :EDIT Thnx to anchint
Solving we get T = 302.2K app
Whats wrong with my method?
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Let us learn to dream, gentlemen, and then perhaps we shall learn the truth.
- August Kekule |
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10 Apr 2008 19:28:48 IST
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seems correct.....I was thinking along anchit's lines...Whats the flaw in that? i doubt whether the increase in volume (0.005l) would bring about so much change..
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Guide to latex:
http://www.goiit.com/posts/list/community-shelf-a-guide-to-latex-48056.htm
JEE and OLYMPIA INFINATUM
http://iit-redefined.theforum.name/index.php
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10 Apr 2008 19:41:29 IST
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Yes, you are right there.
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Will nip in at times to solve problems :)
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10 Apr 2008 19:43:43 IST
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i think conjurer's soln is correct now
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10 Apr 2008 20:30:17 IST
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ok got it
i think my solution is also right cos even if we take into account PdV in the line before eqn 1 its value would be so small that the difference in the answer in the end would be of around 0.25K (i calculated it)
in conjurer's soln,
he has made mistake in taking P2 = 81 ,it is actually 76.5 cm
and now his answer would also come to be the same i think
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10 Apr 2008 21:16:09 IST
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@anchit WTh cant your bear change his dance??? I have been wasting my time watching your bear hoping that he would do some new steps...
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"I a universe of atoms.......an atom in the universe" |
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10 Apr 2008 21:21:56 IST
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lol
anand you really have high hopes !!
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11 Apr 2008 08:22:11 IST
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good work anchit...thanks!
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Guide to latex:
http://www.goiit.com/posts/list/community-shelf-a-guide-to-latex-48056.htm
JEE and OLYMPIA INFINATUM
http://iit-redefined.theforum.name/index.php
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