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Physical Principles in Biology Biology 3550 Fall 2016 Lecture 25 Thermodynamics II: Reversible Processes and Entropy Friday, 28 October c David P. Goldenberg University of Utah [email protected] Important Thermodynamic Concepts and Quantities Introduced so Far Energy: ability to do work, with units of joules or calories. E , internal energy. Directly related to temperature and motion of molecules. ∆E , change in internal energy. ∆E = Efinal − Estart w , work done by the surroundings on the system during a process. q, heat flowing into the system from the surroundings during a process. Adiabatic: without heat flow in or out of the system, q = 0. Isothermal: Without temperature change, ∆T = 0 and ∆E = 0. The first law of thermodynamics: ∆E = q + w Clicker Question #1 Which of the following are true? 1 ∆E < 0 4 q<0 7 w <0 2 ∆E = 0 5 q=0 8 w =0 3 ∆E > 0 6 q>0 9 w >0 Perfect insulation Up to 3 answers accepted. Correct answers are worth 1 2 point. Wrong answers cost Total guaranteed to be ≥ 0. 1 2 point. Clicker Question #2 Weight Which of the following are true? Weight 1 ∆E < 0 4 q<0 7 w <0 2 ∆E = 0 5 q=0 8 w =0 3 ∆E > 0 6 q>0 9 w >0 Up to 3 answers accepted. Correct answers are worth 1 2 point. Wrong answers cost Total guaranteed to be ≥ 0. 1 2 point. Clicker Question #3 Weight Which of the following are true? Weight Thermal Reservoir (Constant T) 1 ∆E < 0 4 q<0 7 w <0 2 ∆E = 0 5 q=0 8 w =0 3 ∆E > 0 6 q>0 9 w >0 Thermal Reservoir (Constant T) Up to 3 answers accepted. Correct answers are worth 1 2 point. Wrong answers cost Total guaranteed to be ≥ 0. 1 2 point. State Functions and Path-dependent Functions State functions of a system depend only on the current state of the system and do not depend on history. Examples: • Temperature, T • Pressure, P • Volume, V • Internal energy, E For any change in a system, the change in a state function depends only on the beginning and ending states. ∆T = Tfinal − Tstart ∆P = Pfinal − Pstart ∆V = Vfinal − Vstart ∆E = Efinal − Estart Work, w , and heat, q, are not state functions. The Maximum-work Path for Gas Expansion Weight Weight Weight Thermal Reservoir Thermal Reservoir Thermal Reservoir Piston is allowed to move upward in infinitesimally small steps. Temperature is never allowed to drop. Pressure drops as gas expands, so less work is done per step. If larger steps are ever taken: • The temperature drops. • The pressure drops more than it would in an infinitesimal step. • Less work is produced. The Minimum-work Path for Gas Compression Weight Weight Weight Thermal Reservoir Thermal Reservoir Thermal Reservoir Piston is pushed down in infinitesimally small steps. Energy is transferred from piston to gas molecules. Temperature is never allowed to increase. Excess energy flows to the reservoir as heat. P increases as gas is compressed, so more work is required per step. If larger steps are ever taken, more work is required. A Reversible Cycle of Compression and Expansion Weight Weight Weight Thermal Reservoir Thermal Reservoir Thermal Reservoir Steps in both directions are infinitesimal. Compression and expansion are exactly the reverse of one another. wcomp = −wexp = −qcomp = qexp For the complete cycle: ∆E = 0 w =0 q=0 Either compression or expansion can be reversed at any point by an infinitesimal force in the opposite direction. Another Kind of State Function w and q are not state functions. BUT, we can define the value of w (or q) for a specific process linking two states to be a change in a state function. We define the work for the reversible (infinitely slow) conversion of one state to the another, wrev , to be the change in state function ∆F . • ∆F is called the change in “free energy.” (Helmholtz free energy) • ∆F is the maximum amount of work that can be obtained from the change in state. • ∆F is “free energy” in the sense that it is energy that is available to do work. • “Free energy” is really the most expensive kind of energy. It’s the kind that we pay for to do work. The heat for a reversible process, qrev is also a state function that we will return to shortly. Calculating the Work for the Reversible Isothermal Expansion of a Gas Weight Weight Weight Thermal Reservoir Thermal Reservoir Thermal Reservoir The work is calculated by integrating force with respect to distance. Z x2 fdx wrev = x1 f is force, and x represents the position of the piston. The force can be expressed in terms of pressure, P and the area that the piston presents to the gas, A: f = −P/A Negative sign indicates that the pressure force is outwards. Calculating the Work for the Reversible Isothermal Expansion of a Gas Weight Weight Weight Thermal Reservoir Thermal Reservoir Thermal Reservoir From the previous slide: Z x2 Z x2 wrev = fdx = − P · Adx x1 x1 For each small increment in x, there is a corresponding increment in volume, dV . dV = dxA dx = dV /A Calculating the Work for the Reversible Isothermal Expansion of a Gas Weight Weight Weight Thermal Reservoir Thermal Reservoir Thermal Reservoir Substituting dV /A for dx in the work integral: Z x2 Z V2 PdV P · Adx = − wrev = − x1 V1 (Recall that the product PV has the units of energy or work!) From the ideal gas law, P = nRT /V . Substituting: Z V2 Z V2 nRT dV wrev = − dV = −nRT V V1 V1 V Calculating the Work for the Reversible Isothermal Expansion of a Gas Weight Weight Weight Thermal Reservoir Thermal Reservoir From the previous slide: Z V2 dV wrev = −nRT V1 V From calculus: V2 V2 wrev = −nRT ln(V ) = −nRT ln V1 V1 wrev depends on n, T , V1 and V2 . wrev represents a change in a state function. Thermal Reservoir Calculating the Work for the Reversible Isothermal Expansion of a Gas Weight Weight Weight Thermal Reservoir Thermal Reservoir Thermal Reservoir Since the number of moles remains constant, the change in volume can also be expressed in terms of a change in concentration: C1 V1 = C2 V2 V2 /V1 = C1 /C2 Substituting into the result from the previous slide: C1 wrev = −nRT ln C2 qrev for Isothermal Expansion For an isothermal process: ∆E = q + w = 0 q = −w For reversible isothermal expansion: C2 qrev = −wrev = −nRT ln C1 Sign check: For expansion, C2 < C1 and qrev > 0, and the flow is into the system, as we expect. For any two states, there is, in principle, a reversible (infinitely slow) process separating them, and we can define qrev and wrev . Reconsider the adiabatic expansion without work q=0 Insulation prevents heat flow w =0 No work is done ∆E = q + w = 0 Perfect insulation Consistent with no temperature change. We know that something has been lost, because restoring the original condition would require work. We suspect that this has to do with the loss of order, or increase in entropy. But, what is entropy? How do we give it a number? The “Classical” Definition of Entropy, S For two states for which temperature remains constant during the reversible process of converting one to the other: qrev T For two states separated by a reversible process for which the temperature does not stay constant: ∆S = Z T2 ∆S = T1 qrev dT T Units for entropy: energy/temperature, J/K. What does this have to do with randomness? Entropy Change for Isothermal Expansion of a Gas From before, qrev : qrev = −wrev V2 = nRT ln V1 C1 = nRT ln C2 Entropy: qrev V2 C1 ∆S = = nR ln = nR ln T V1 C2 ∆S is positive if volume increases (or concentration decreases). ∆S does not depend on temperature (in this case). The Statistical Definition of Entropy S = k log W k = Boltzmann’s constant, with correct units for entropy (J/K). S = k ln Ω What is W (or Ω)? Ludwig Boltzmann, 1844–1906 Tombstone in Vienna, Austria For a given state, Ω is the number of equally probable ways to arrange the components making up that state. The different arrangements are called microstates. There is no proof! We believe it because it works.