use the Maxwell relations to express thederivatives (a) ( S/ V)T and ( V/S)p (b)
ID: 687648 • Letter: U
Question
use the Maxwell relations to express thederivatives(a) ( S/ V)T and ( V/S)p (b) ( p/S)V and ( V/S)p in terms of the heat capacities, the expansioncoefficient , and the isothermal compressibility T. use the Maxwell relations to express thederivatives
(a) ( S/ V)T and ( V/S)p (b) ( p/S)V and ( V/S)p in terms of the heat capacities, the expansioncoefficient , and the isothermal compressibility T. in terms of the heat capacities, the expansioncoefficient , and the isothermal compressibility T.
Explanation / Answer
a) dA = -SdT - PdV => dS/dV|T = dP/dT|V dP/dT|V*(dT/dV)|P *(dV/dP)|T = -1 dP/dT|V*(1/PV) *(-KT*V) = -1 dP/dT|V*(1/P) *(KT) = 1 dP/dT|V = dS/dV|T = P/KT ------------------------------ dV/dS ??? dH = TdS + VdP dV/dS|P = (dT/dP)|S dT/dP|S *(dP/dS)|T*(dS/dT)|P = -1 dT/dP|S *(dP/dS)|T*(Cp/T) = -1 dG = -SdT + VdP dS/dP |T = dV/dT|P = V*P dT/dP|S *V*P*(Cp/T) = -1 dV/dS|P = dT/dP|S = -T/(Cp*P*V) ============== b) dP/dS|V dU = TdS - PdV dP/dS|V = -dT/dV|S dT/dV|S*(dV/dS|T)*(dS/dT)|V = -1 dA = -SdT - PdV => dS/dV|T = dP/dT|V dT/dV|S*(dV/dS|T)*(dS/dT)|V = -1 dT/dV|S*(dP/dT|V)*(Cv/T) = -1 dP/dT|V*(dT/dV)|P*(dV/dP)|T = -1 dP/dT|V * (1/P*V)(-KT*V) = -1 dP/dT|V = P/KT dT/dV|S = -T*KT/(pCV) dP/dS|V = -dT/dV|S =T*KT/(pCV) ===================== dV/dS|P this was already done in 2nd part of a)
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.