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Ch. 5 – 2 nd Law of Thermodynamics. Entropy The 1 st Law is a statement of energy conservation : implies that we cannot create or destroy energy. However, it does not place any limits on how energy can be transformed from one form to another;
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Ch. 5 – 2nd Law of Thermodynamics • Entropy • The 1st Law is a statement of energy conservation: implies that we cannot create or destroy energy. • However, it does not place any limits on how energy can be transformed from one form to another; • According to 1st Law: heat can be transformed into work, work into heat, work can be done at the expense of internal energy, etc. • If other laws do not exist, the 1st Law would allow certain phenomena to happen that will never happen in reality.
Ch. 5 – 2nd Law of Thermodynamics • Entropy • Natural thermodynamic processes are, to one degree or another, irreversible; Such processes move toward equilibrium. • Reverse processes are not excluded by 1st Law. Imagine these examples: • Compression of gas under no external pressure • Heat flowing from a colder body to a warmer one • We know that these processes are impossible in nature. • Since they are not excluded by 1st Law, we need something else. • The 2nd Law of Thermodynamics takes care of this.
Ch. 5 – 2nd Law of Thermodynamics • Entropy • Depending on external conditions, natural processes can produce opposite changes in a system. • Irreversibility of such processes can be reduced by modifying their paths. • Attempt to change path so difference between actual values of state variables and values corresponding to equilibrium is reduced at all stages; if we continue doing this indefinitely, we tend to a common limit for processes producing either one change or its opposite. • This limit is called a reversible process, therefore a reversible processis an ideal limit
Ch. 5 – 2nd Law of Thermodynamics • Entropy • Limit is a reversible process. • Ideal limit – cannot be realized • Can approximate, however • Can be defined as • A series of states that differ infinitesimally from equilibrium • States succeed each other infinitely slowly • Variables change in continuous way • Underlying assumption – the rate of the process tends to zero as equilibrium approached
Ch. 5 – 2nd Law of Thermodynamics • Entropy • We can reframe our earlier examples to find a reversible analogue examples. • Expansion of gas with pressure p against external pressure p – Δp • Heat conduction along an infinitesimal temperature gradient • By reversing the sign of infinitesimal difference from equilibrium, reversible process of opposite sense occurs (replace dp by –dp, or dT/dx by –dT/dx.) • Key idea, slow changes departing very little from equilibrium.
Ch. 5 – 2nd Law of Thermodynamics • Entropy • For every reversible process there is a positive integrating factor 1/of the differential expression Q, which is only dependent on the thermal state and which is equal for all thermodynamic systems. • This defines a state function S, the entropy, according to the exact differential,
Ch. 5 – 2nd Law of Thermodynamics • Entropy • The following is found to be true for irreversibleprocesses • We can combine the two expressions to generalize for any process • The inequality and equality symbols correspond to irreversible and reversible processes, respectively. • MATHEMATICAL EXPRESSION OF 2nd LAW – also called as the supreme law of nature.
Ch. 5 – 2nd Law of Thermodynamics • Carnot Cycle (thermal engine: receives from a source an amount of heat, and part of it is transformed into work) • All processes are reversible. • Contains 2 isotherms, T1 and T2 < T1, 2 adiabats 2 < 1. • Steps are • Isothermal expansion T1 = const • Adiabatic expansion 1 = const • Isothermal compression T2 = const • Adiabatic compression 2 = const
Ch. 5 – 2nd Law of Thermodynamics • Carnot Cycle (ideal gas) • Calculate work done and heat absorbed. • Step 1 (isothermal; ∆T=0) Since V2>V1, W12,Q12>0 -work was done by the gas and heat was absorbed by the gas from the heat source at T1
Ch. 5 – 2nd Law of Thermodynamics • Carnot Cycle (ideal gas) • Step 2 (adiabatic; Q23 = 0) • Since V3>V2, W23>0 • -work was done by the gas
Ch. 5 – 2nd Law of Thermodynamics • Carnot Cycle (ideal gas) • Step 3 (isothermal; ∆T=0) Since V4<V3, W34,Q34<0 -work is done on the gas and heat is given away by the gas to the source at T2
Ch. 5 – 2nd Law of Thermodynamics • Carnot Cycle (ideal gas) • Step 4 (adiabatic; Q41 = 0) • Since V1<V4, W41<0 • -work is done on the gas • Total work done is (shaded)