Consider this: this problem appeared in our CYL110 (a 2nd semester chemistry course) quiz...half the CS department couldn't get it right!
This gas undergoes a reversible process s depicted in the P-V curve...the center of the circle shown is (10,10) and its radius is 5;
You know whats coming...find the maximum temperature of the gas!
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ReplyDeleteoh god...too much algebra involved!!!
ReplyDeletesrry....made a blunder...is it at (10+(5/root2), 10+5/root2))
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ReplyDeletevaibhav, your approach was nearly the same as what i applied...though you could have used some other curve instead of the 'line'..
ReplyDeletewhat could be this 'some other curve'???
anuj bhaiya ,can u please give ur approach
ReplyDeleteyeah got same ans as vaibhav,
ReplyDeletesolved it like a trigo question,realised something obvious on seeing the ans
oh shit...was playing with algebra
ReplyDeleteparametric equation proved much, much easier..!!
yea the answer was obvious already
getting nRT as 100 + 12.5 + (50sqrt[2])
Bhaiya...i cud not get d curve u were talking about...
ReplyDeletehi Vaibhav...you approach was cool, but i used the hyperbola PV=const...kind of the obvious curve for the isotherm...
ReplyDeleteas the distance of the hyperbola increases, temperature increases...then the answer follows from Symmetry!...
and asking for the maxm. temp was in a way incorrect...should have specified the no. of moles...
hm..
ReplyDeleteeqn is of a circle (x-10)^2 + (y-10)^2 = 25
for max temp
xy shud be max which at x = y (by am>gm)
ans is at P=V=10+(5/root2)
hi ankush...cool use of THE inequality!!
ReplyDeleteSimple AM>GM was enough.. it would be when P=V and both are max so this gives 10+5/root2 P and same V.
ReplyDeletesame approach as ankush