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[Post New]10 Apr 2008 19:16:58 IST
    Subject: Re:Thermodynamics/Fluids question
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anchitsaini (4315)

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Olaaa!! Perrrfect answer. 789  [974 rates]
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mbox{ increase in pressure }= dP = hdg \ \ = 5*10^{-3} * 13.6 * 10^3 * 10 = 136 *5Pa \ \ mbox { now }PV = nRT \ \ mbox{ on differentiating } PdV + VdP = nRdT \ \ mbox{as dV = 0 }\ \ VdP = nR dT ....1\ \ mbox{Also at 300K , volume occupied by 1 mole of gas }= rac{22.4*300}{273} \ \ mbox{hence number of moles of gas occupying 7 l of gas at 300K } = rac{7*273}{22.4*300}\ \=0.28 = n \ \ mbox{putting this value in eqn 1, }\ \7*10^{-3}*{136*5} = 0.28*8.314*dT \ \ dT = 2K\ \ T = 302K
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