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can i get bit mesra with 110 in jee main and 89% marks in cbse board.i belong to general category. can i get bit mesra with 110 in jee main and 89% marks in cbse board.i belong to general category.
Start with the definition of heat capacity at constant pressureCp = (?H/?T)p ((..)p indicates derivation at constant pressure)Substitute the enthalpy by its definitionH = E + pVTherefore:Cp = (?E/?T)p + (?(pV)/?T)p = (?E/?T)p + p×(?V/?T)pBecausedE = (?E/?V)t dV + (?E/?T)v dT = (?E/?V)t dV + Cv dTThe partial derivative of E with respect to temperature at constant pressure is:(?E/?T)p = (?E/?V)t (?V/?T)p + CvJoin both expression together:Cp - Cv = (?E/?V)t (?V/?T)p + p×(?V/?T)p = [(?E/?V)t + p]×(?V/?T)pq.e.d.For further calculations the is a useful relation to evaluate the partial derivative of the energy from an equation of state.As shown in my answer to your last question (see link):(?E/?V)t = T×(?p/?T)v - pFor an ideal gaspV = RT <=> p = (RT)/V <=> V = (RT)/p(?E/?V)t = T×R/V - p = 0(?V/?T)p = R/pCp - Cv = [p + 0] × R/p = RFor a Van der Waals gas(p + a/V²) = (RT)/(V-b)As shown before (see link):(?E/?V)t = a/V²The partial derivative of the volume you get by implicit differentiation(p + a/V²) - (RT)/(V-b) = 0-R/(V-b) +[-2a/V³ + (RT)/(V-b)²] (?V/?T)p = 0=>(?V/?T)p = [R/(V-b)] / [(RT)/(V-b)² -2a/V³]Cp - Cv = [p+a/V²] × [R/(V-b)] / [(RT)/(V-b)² -2a/V³]= [(RT)/(V-b)] × [R/(V-b)] / [(RT)/(V-b)² -2a/V³]= R / (1 - [2a×(V-b)²]/[RTV³])Thanks & RegardsRinkoo GuptaAskIITians Faculty
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