Consider ‘n’ moles of an ideal gas contained in a cylinder fitted with a frictionless piston. If the piston is fixed and the gas is heated, its volume remains constant and all the heat supplied goes to increase the internal energy of the molecules due to which the temperature of the gas increases. If DQv is the amount of heat supplied and DT is the rise in temperature then, DQv = n CvD T For latest information , free computer courses and high impact notes visit : citycollegiate The pressure of the gas increases during this process, but no work is done because the volume is kept constant. Hence D W = 0. applying first law of thermodynamics Heat supplied = Increase in internal energy + Work done DQv = D U + 0 DQv = D U OR nCvD T = D U If the piston is free to move, the gas may be allowed to expand at a constant pressure. Let the amount of heat supplied is now is DQp. The addition of heat causes two changes in the system: Increase in internal energy Work done against external pressure According to the first law of thermodynamics: DQ = DU + DW {But DW = PDV} DQP = DU + PDV Since DQp = nCpDT and DU = nCvDT , therefore, nCpDT = nCvDT + PDV....................(1) We know that PV = nRT At T1 Kelvin: PV1 = nRT1 .....(a) At T2 Kelvin: PV2 = nRT2.....(b) Subtracting (a) from (b) PV2 - PV1= nRT2 - nRT1 P(V2 - V1)= nR(T2 - T1) {(V2 - V1) = DV and (T2 - T1) = DT } PDV = nRDT Putting the value of PDV in equation (1) nCpDT = nCvDT + nRDT nCpDT = nDT(Cv + R) Cp = (Cv + R) Cp - Cv = R For latest information , free
Posted on: Tue, 25 Mar 2014 17:21:29 +0000
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