作者: kramnik (progressive) 看板: Physics 标题: Re: [问题] 关於热力学第二定律 时间: Fri Jan 8 21:35:54 2010 ※ 引述《marchant (蚂蚁)》之铭言： : 克劳休斯不等试 : 中的dQ是输出热还是输入热? 对於一隔离系统..我们将其分为反应系统(s)及环境(e)两部份.. clausius theorem若采用标准型式 ∫s [δQ/T(s)] ≦ 0 则其中的δQ为环境(e)给予反应系统(s)的热量 ================================================================ 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 上三者为等效论述 ==================================================================== dS ≧δq/T => clausius theorem 之推导过程 根据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 .............(##) (#)(##)合并即得 clausius theorem --

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