《物理化学 Physical Chemistry》课程教学资源（英文版）The Second Law（The Concepts）

The second law The Concepts Chemistry 331

Chemistry 331 The Second Law The Concepts

The Laws of Thermodynamics First law You cant get something for nothing Second law You can t even break even Chemistry 331

Chemistry 331 The Laws of Thermodynamics • First Law You can’t get something for nothing. • Second Law You can’t even break even

Heat engines a heat engine converts heat into work Th=temperature ofheat source Engine c=temperature of heat sink gh= heat supplied 」qn qe heat released W=work produced Chemistry 331

Chemistry 331 Heat Engines • A heat engine converts heat into work. • Th = temperature of heat source • Tc = temperature of heat sink • qh = heat supplied • qc = heat released • w = work produced Th Tc Engine qh qc w

efficiency of heat engines An engine operates in You can iget a cycle:. AU=O something for and w=gh-lqcl nothing. efficiency is given b 8=w /qr q cl / gh q>o implies a< 1 You can t even breakeven Chemistry 331

Chemistry 331 Efficiency of Heat Engines • An engine operates in a cycle; \ DU = 0 and |w| = qh - |qc | • efficiency is given by e = |w| / qh \ e = 1 - |qc | / qh • |qc |>0 implies e < 1 You can’t get something for nothing. You can’t even break even

The Second law(Kelvin) No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work In other words, w<gh and the efficiency of a heat engine is less than 100% Chemistry 331

Chemistry 331 The Second Law (Kelvin) • No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work. • In other words, |w| < qh and the efficiency of a heat engine is less than 100%

Propositions For a reversible cycle, absorbing q, atT and g2 at T2, the ratio q: q2 depends only on Tand t2 2. This ratio may be written: q q2=t,/ T2 3. Th he entropy, S, defined by ds=(dq/t) rev S a function of state 4. In an isolated system, As is zero for a reversible process, and positive for a spontaneous one Chemistry 331

Chemistry 331 Propositions 1. For a reversible cycle, absorbing q1 at T1 and q2 at T2 , the ratio q1 :q2 depends only on T1 and T2 . 2. This ratio may be written: q1 /q2 = T1 /T2 3. The entropy, S, defined by dS=(dq/T)rev is a function of state. 4. In an isolated system, DS is zero for a reversible process, and positive for a spontaneous one

The Entropy function The entropy s is defined by reversible state B y ds= dg/t' ds-dqre T dS≠ dqirrey/ irreversible 4S≠d/T state A Chemistry 331

Chemistry 331 • The entropy S is defined by dS = dqrev/T • dS  dqirrev/T The Entropy Function reversible: dS = dq/T irreversible: dS  dq/T state A state B

The Second Law(clausius) The entropy s is a function of state ds >dq / T(clausius inequality o ds= dq /t for a reversible change o ds> dg/t for an irreversible change Corollary For an isolated system ds=0 for a reversible change ds>o for an irreversible change Chemistry 331

Chemistry 331 The Second Law (Clausius) • The entropy S is a function of state. • dS  dq / T (Clausius inequality) • dS = dq / T for a reversible change • dS > dq / T for an irreversible change • Corollary: For an isolated system • dS = 0 for a reversible change • dS > 0 for an irreversible change

Carnot cycle isothermal: 100 K adiabatic isothermal: 200 K diabatic V/L Chemistry 331

Chemistry 331 Carnot Cycle 0 1 2 3 0 10 20 30 40 50 V / L p / atm isothermal: 100 K adiabatic isothermal: 200 K adiabatic

Carnot efficiency △S=△S +△S≥0 engine but△S =0 engine △S surr (qh/T)+(q/T)≥0 8 E≤1-(T/T Chemistry 331

Chemistry 331 Carnot Efficiency DS = DSengine + DSsurr  0 but DSengine = 0 \ DSsurr = - (qh / Th ) + (qc / Tc )  0 \ qc / qh  Tc / Th e = |w| / qh = 1 - |qc | / qh \ e  1 - (Tc / Th )