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Solution: d omega = force time distance but p = f/Delta Total workdone = integral^Final_initial dw = w = integral^V_final_V_initial P dv when the condition ds = 0, PV_initial = n_initial RT PV_initial = n_equal RT P(V_final = V_initial) = (n_final = n_initial) RT P Delta V - Delta nqRT, Delta H = Delta + P Delta V Delta h = Delta V + Deltan gRt When ds = 0, (-dv) + dg_B + dw = 0 or dv - dg_B - dw = 0 in equation (reversiable condition) DeltaV = 0, So g = w integral^V_2 _ V_1 P dv (Therefore q = Deltav + w = w) For n mole of gas, we have PV = nRT (or) P = nRT/V on substituting we get, w = integral^V2_V1 nRt dv/V = nRT logV_1/V_2 = nRT log P_1/P_ 2 On changing the base of logarithm from e to log, we get = 2.303 nRT log V_2/V_1 = 2.303nRT log V_2/v_1
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