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One mole of a mono- atomic ideal gas undergoes a quasi- static process, which is depicted by a straight line joining points (V_(0)T_(0)) " and " (2V_(0) , 3T_(0)) in a V - T diagram . What is the value of the heat capacity of the gas at the point (V_(0) , T_(0)) ? |
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Answer» R ` :. ` Work done , `dW = PdV`CHANGE in internal energy is, `Delta U = nC_(V) Delta T` where `C_(V)` = specific heat at constant volume. Heat given in process is , `Delta Q = nCDelta T `,  where n is number of moles of gas , C is molar heat capacity. ACCORDING to graph, temperature is increasing with increase in volume. It means that energy is used both in increasing internal energy and work done. According to `1^(st)` law of thermodynamics ` PdV + nC_(V)dT = nCdT` or ` P/n (DV)/(dT) + C_(V) = C` .....(i) At `(V_(0) , T_(0)) (dV)/(dT) = V_(0)/(2T_(0)) ` ....(ii) Again , `P_(0)V_(0) = nRT_(0) rArr P_(0)/n = (RT_(0))/V_(0) ` ....(iii) From (i), (ii) and (iii) , `C = (RT_(0))/V_(0) V_(0)/(2T_(0)) + 3/2 R = 2 R [ :. " for monoatomic gas , " C_(V) = 3/2 R]` |
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