98專利師普化,物化討論 - 專利

By Wallis
at 2010-05-17T22:48
at 2010-05-17T22:48
Table of Contents
剛好路過,解物化第一題^^
※ 引述《Xkang (日文專利工程師)》之銘言:
: <物化>
: 一、有一莫爾(mole)氣體,其狀態方程式為
: pV = RT + B(T) p ,B(T) ≠ 0
: 其中,p、V、T 為氣體的壓力、體積及溫度,R 為氣體常數,B(T)只是T 的函數。
: 在固定溫度之下,發現氣體的焓(enthalpy)與體積無關。試求B 與T 之間的函數
: 關係。(15 分)
1. 已知 At const. T, enthalpy(H) 非 V 函數
H對V偏微 定T下=0
H=H(S,T) dH=TdS+VdP----(a)
(a) 對V偏微, at const. T
得 @ const.T (dH/dV)=T(dS/dV)+V(dP/dV)=0--------(b)
打不出偏微分符號XD
2. 由Maxwell eq. 將(dS/dV) sub. T 換為 (dP/dT) sub. V
3. PV=RT + B --> P= RT/V + B/V
解P對T偏微@ const. V , and P對V偏微@ const.T
代入(b)
4. 得dB/B = dT/T ln B = ln T + A' --> B = AT
請參考~
--
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