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Chemistry · Ch 7 — Thermodynamics

Various Statements of the Second Law

7.10.1

Various Statements of the Second Law

Entropy. The second law introduces a new state function, entropy (SS), as a measure of a system's molecular disorder (randomness). Its rigorous thermodynamic definition concerns the CHANGE in entropy resulting from a process:

dS=dqrevTdS = \frac{dq_{rev}}{T}

Entropy statement of the second law: the entropy of an isolated system increases during a spontaneous process. For an irreversible (spontaneous) process, such as the spontaneous expansion of a gas, ΔStotal>0\Delta S_{total}>0, and more precisely

ΔStotal>ΔSsystem+ΔSsurrounding\Delta S_{total} > \Delta S_{system}+\Delta S_{surrounding}

For a REVERSIBLE process, such as ice melting at its melting point under equilibrium conditions, ΔSsystem=−ΔSsurrounding\Delta S_{system}=-\Delta S_{surrounding}, so the two exactly cancel and ΔStotal=0\Delta S_{total}=0.

Kelvin-Planck statement: it is impossible to construct a machine that absorbs heat from a hot source and converts it COMPLETELY into work by a cyclic process, without transferring some part of that heat to a cold sink. This is exactly why even an ideal, perfectly frictionless engine can never convert 100% of the heat it absorbs into useful work. Carnot's analysis of heat engines showed that the MAXIMUM efficiency of a reversibly-operating heat engine depends only on the two temperatures it operates between:

η=qh−qcqh=1−qcqh(7.27)\eta = \frac{q_h-q_c}{q_h} = 1-\frac{q_c}{q_h}\qquad(7.27)

where qhq_h is heat absorbed from the hot reservoir and qcq_c is heat rejected to the cold reservoir. For a reversible cyclic process, ΔStotal=ΔSsystem+ΔSsurroundings=0\Delta S_{total}=\Delta S_{system}+\Delta S_{surroundings}=0, so ΔSsystem=−ΔSsurroundings\Delta S_{system}=-\Delta S_{surroundings}, and expressing each in terms of q/Tq/T gives qhTh=qcTc\dfrac{q_h}{T_h}=\dfrac{q_c}{T_c} (7.28); substituting this ratio back into (7.27) gives

η=1−TcTh(7.29)\eta = 1-\frac{T_c}{T_h}\qquad(7.29)

Since Th≫TcT_h\gg T_c in any real engine, η\eta is always less than 1 -- perfect (100%) conversion of heat to work is thermodynamically forbidden. Expressed as a percentage:

% Efficiency=(1−TcTh)×100\%\ \text{Efficiency} = \left(1-\frac{T_c}{T_h}\right)\times100

Worked example (Problem 7.10). An automobile engine burns petrol at 816∘816^\circC, with surroundings at 21∘21^\circC. Converting to Kelvin: Th=816+273=1089T_h=816+273=1089 K, Tc=21+273=294T_c=21+273=294 K. …

Misc 7.10Problem 7.10 -- maximum efficiency of an automobile engine

Worked out. An automobile engine burns petrol at Th=816+273=1089T_h=816+273=1089 K, with surroundings at Tc=21+273=294T_c=21+273=294 K. %Efficiency=(Th−TcTh)×100=(1089−2941089)×100≈73%\%\text{Efficiency}=\left(\dfrac{T_h-T_c}{T_h}\right)\times100=\left(\dfrac{1089-294}{1089}\right)\times100\approx73\%. …

Misc evaluate-yourself-5Evaluate Yourself 5 -- efficiency of an engine between 127$^\circ$C and 47$^\circ$C

Worked out. Book's practice box (no printed solution): an engine operates between 127∘127^\circC (Th=400T_h=400 K) and 47∘47^\circC (Tc=320T_c=320 K) with no frictional losses; find the percentage efficiency. Working it through: %Efficiency=(1−320400)×100=20%\%\text{Efficiency}=\left(1-\dfrac{320}{400}\right)\times100=20\% (own solution, not printed in the textbook). …