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![[Post New]](/templates/default/images/icon_minipost_new.gif) 20 Feb 2008 21:55:10 IST
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a cylinder containing oxygen (gamma = 1.4) is closed at the top by a 50 kg frictionless piston.the area of cross section is 100 cm^2. atm. pressure = 100Kpa.
g= 10 m/s^2...the cylinder is heated slowly for some time.find the amount of heat supplied to the gas if the piston moves out through a distance of 20 cm.
(hcv - vol 2 pg 77 qn. 3)
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20 Feb 2008 22:25:58 IST
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pl. post the answer
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kaushik krishna .R
bits pilani
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21 Feb 2008 00:27:08 IST
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(edited)
external pressure = 105 Pa
weight = 50(10) = 500 N
area of cross section = 10-2 sq.m
pressure due to weight of piston = 500/10-2 = .5 * 105
total pressure = 1.5 * 105 Pa .............(1)
change in volume = 10-2 * .2
Work done = P V
= 1.5 * 105 * 10-2 * .2
= 300J
But P V = nRT
nT = 36.08...........(2)
CP = R /(-1) ................(u can easily derive using Cv = fR/2)
CP = 8.314(1.4)/(1.4-1)
CP = 29.1J ......................(3)
Q = nTCp
= 36.08*29.1
= 1050J
plz correct me if wrong at any point.....
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21 Feb 2008 00:53:12 IST
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heyy but he 's given the answer as 1050 J
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kaushik krishna .R
bits pilani
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21 Feb 2008 00:56:08 IST
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howzz w = nCv t
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kaushik krishna .R
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21 Feb 2008 01:01:28 IST
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it is isobaric
pv = nRt
v = 100 x 20 x 10^ -6
from this we can find nt.....
u = nCvt ...we know nt and Cv ......hence we can find the answer.......anyways thanks mate
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kaushik krishna .R
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21 Feb 2008 01:05:04 IST
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its so simple mate...
we found nt naa......
nt = 300
nt Xcp = q
300x3.5 = 1050 J
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kaushik krishna .R
bits pilani
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21 Feb 2008 01:30:26 IST
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perfect!
but im not able 2 follo nythin
since it is possible 2 c many crosses in ur ans!
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VARSHA KRISHNAN....
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IIT is always a word which rises E thru' d body. But 2 achieve it U hav 2 drain out d entire E out of the body....
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21 Feb 2008 09:10:46 IST
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those crosses imply multiplication symbol
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kaushik krishna .R
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