Lect+16+08

# Lect+16+08 - Lecture 16 Second LAW Chapter 13 Sections 13.1...

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Lecture 16 Second LAW Chapter 13 Sections 13.1, 13.2, 13.5 13.6, 13.7 To calculate the entropy change ( S) between state 1 and state 2 we must construct a reversible path (at equilibrium) between state 1 and state 2 Entropy changes Isothermal Changing T S = dq rev /T i f

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Calculating Entropy Changes Use the thermodynamic definition: dS = dq rev /T 1. Define clearly the initial and final states. 1. Find a reversible path linking those states . 1. If T = constant, (isothermal process): calculate q rev for the reversible process. 2. If T is not constant, you must integrate along the reversible path: S = dq rev T state 1 state 2
Plotting thermodynamic States and Processes Thermodynamic states can be plotted as functions of the thermodynamic variables - P, V, T, and the number of moles of each component. The thermodynamic state of the system is defined by a specified set of thermodynamic variables. Any curve connecting two thermodynamic states is called a process . P V n = const T . State 1 . State 2 three different processes connecting State 1 and State 2 One path between two states is the reversible path . The other paths are irreversible . On the reversible path , the system remains arbitrarily close to thermodynamic equilibrium .

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Isothermal Entropy changes Isothermal Reversible process Isothermal Expansion of Ideal Gas E = 0 because dT=0 So along reversible path q rev = -w rev and w rev = - PdV = - nRT dV/V 2200 S = nR ln(V f /V i ) S = dq rev /T i f =1/T dq rev =1/T dw rev =1/T nRT dV/V i f i f i f S = dq rev /T i f =1/T dq rev = q rev /T i f
Isothermal Expansion of an Ideal Gas n 1 = 0.25 mole T 1 = 25°C V 1 = 2.00 L P 1 = 3.06 atm n 1 = 0.25 mole T 1 = 25°C V 1 = 10.00 L P 1 = 0.611 atm S = ?

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