作者: kramnik (progressive) 看板: Physics 標題: Re: [問題] 關於熱力學第二定律 時間: Fri Jan 8 21:35:54 2010 ※ 引述《marchant (螞蟻)》之銘言： : 克勞休斯不等試 : 中的dQ是輸出熱還是輸入熱? 對於一隔離系統..我們將其分為反應系統(s)及環境(e)兩部份.. clausius theorem若採用標準型式 ∫s [δQ/T(s)] ≦ 0 則其中的δQ為環境(e)給予反應系統(s)的熱量 ================================================================ clausius theorem ∫circle [δQ(e→s)/T(s)] ≦ 0 (a)If any part of the cyclic process is irreversible (spontaneous), the inequality applied and the cyclic integral is negative. (b)If the cyclic process is reversible, the equality applied. (c)It is impossible for the cyclic integral to be greater than zero. =================================================================== second law of thermodynamic dS ≧δq/T ">"成立於 spontaneous or irreversible process. "="成立於 reversible process. ================================================================== Carnot's theorem <=> clausius theorem <=> dS ≧δq/T 上三者為等效論述 ==================================================================== dS ≧δq/T => clausius theorem 之推導過程 根據second law of thermodynamic dS = δq(r)/T ...............(a) dS > δq(irr)/T ...............(b) ∫1→2→1 δq(r)/T = 0 ........(c) 由(a)知 0 = ∫1→2 δq(r)/T - ∫1→2 dS 0 = ∫1→2 δq(r)/T - ∫1→2 δq(r)/T 由(c)知 0 = ∫1→2 δq(irr)/T + ∫2→1 δq(r)/T ...............(#) 由(b)知 0 > ∫1→2 δq(irr)/T - ∫1→2 dS 由(a)知 0 > ∫1→2 δq(irr)/T - ∫1→2 δq(r)/T 由(c)知 0 > ∫1→2 δq(irr)/T + ∫2→1 δq(r)/T .............(##) (#)(##)合併即得 clausius theorem --

※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 118.168.81.226 ※ 編輯: kramnik 來自: 118.168.81.226 (01/09 17:07)