Gibbs-Helmholtz equation 公式(3)(A01 △4-△U7 的导出 aT T 根据基本公式dA=-S7-pd OA a(△A S △S aT aT 根据定义式A=U-TS 在7温度时△4=△U/-T△S则△S= △A-△U T (△A)1△4-△C 所以 aT 4上一内容下一内容◇回主目录 返回 2021/223
上一内容 下一内容 回主目录 返回 2021/2/23 Gibbs-Helmholtz equation 根据基本公式 d d d A S T p V = − − ( ) ( ) [ ] V V A A S S T T = − = − 根据定义式 A U TS = − 在T温度时 = − A U T S 所以 ( ) [ ]V A A U T T − = 公式 的导出 ( ) (3) [ ] V A A U T T − = A U S T − 则 − =
Gibbs-Helmholtz equation △4 a( 公式(4)[ 7s、△U 的导出 在公式(3)两边各乘得 1ro(△A)1_△A-△U T aT 移项得 O(△A)1△4△U T aT △A 等式左边就是()对微商的结果,则 ( △ △A OT 移项积分得∫d(y)=「 △U dT T △A 知道U,C与T的关系式,就可从求得y的值。 4上一内容下一内容令回主目录 返回 2021/223
上一内容 下一内容 回主目录 返回 2021/2/23 在公式(3)两边各乘 得 1 T Gibbs-Helmholtz equation 2 1 ( ) [ ]V A A U T T T − = 2 ( ) [ ]V A T U T T = − 移项得 2 2 1 ( ) [ ]V A A U T T T T − = − 等式左边就是 ( ) 对T微商的结果，则 A T 公式 2 的导出 ( ) (4) [ ]V A T U T T = − 移项积分得 2 d( ) d V A U T T T = − 知道 U C, V 与T的关系式，就可从 求得 的值。 1 A T 2 A T
Derive and use the Clapeyron equation Suppose both p and t are made to change by an infinitesimal amount, but in such a way that the two phase remain in equilibrium The new values of the chemical potentials remain equal, and so the changes du(a, P,T)and du(β,PT) must be equal We know how to express an infinitesimal change in u in terms of changes in the pressure and temperature and so 4上一内容下一内容令回主目录 返回 2021/223
上一内容 下一内容 回主目录 返回 2021/2/23 Derive and use the Clapeyron equation Suppose both p and T are made to change by an infinitesimal amount,but in such a way that the two phase remain in equilibrium. The new values of the chemical potentials remain equal, and so the changes d(,P,T) and d(,P,T) must be equal. We know how to express an infinitesimal change in in terms of changes in the pressure and temperature and so
du(a,P,T)=du(β,P,T) implies -Sn(a)dT+Vn(a)dP=-Sn(β)dT+Vn(β)dP where Sm(a) and Sm()are the molar entropies of the two phases and vm(a) and vm(b)their molar volumes Rearranging this gives iVm(o-vmB)dP=Sm(a)-Sm(B))dT 4上一内容下一内容令回主目录 返回 2021/223
上一内容 下一内容 回主目录 返回 2021/2/23 d (,P,T) = d (,P,T)))) implies - Sm()dT + Vm()dP = - Sm()dT + Vm()dP where Sm() and Sm() are the molar entropies of the two phases and Vm() and Vm() their molar volumes. Rearranging this gives {Vm()- Vm()}dP = {Sm() - Sm() }dT
This important result is called the Clapeyron equation It is a thermodynamic result, it is exact, and it applies to any phase change of pure materials dP/dT=△S/△V The△ s for a phase transition under equilibrium conditions is △S=△H/T so that dP△H m CTT△V m 4上一内容下一内容令回主目录 返回 2021/223
上一内容 下一内容 回主目录 返回 2021/2/23 This important result is called the Clapeyron equation. It is a thermodynamic result, it is exact, and it applies to any phase change of pure materials. dP / dT = △Sm / △Vm △Sm = △Hm / T m m T V H = dT dP The S for a phase transition under equilibrium conditions is so that