Chapter 3

# Asaresultofthedifferenceintheirinitial

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Unformatted text preview: ‐ The Second Law of Thermodynamics  98    Entropy Changes for the System during Specific Reversible Processes:    1)  Isothermal Reversible Expansion or Compression of an Ideal Gas  dU = δw + δq = CVdT = 0  δq = ‐δw = PextdV = P dV = nRT (dV/V)  dS = δq / T =  nR (dV/V)  Note dU = 0 because the process is isothermal for an ideal gas.  Note Pext = P because the process is reversible.  Note dS = dδ / T because the process is reversible.  Note when a process is reversible, it is mechanically reversible (Pext = P) and  it is thermally reversible (Tsurr = T)  Therefore, ΔS = nR ln (Vf / Vi)  2)  Reversible Adiabatic Process (all materials)  If a process is reversible, then dS = δq / T  If the process is adiabatic then δq = 0. Therefore, for a reversible adiabatic  process, dS = 0.  One says that a reversible adiabatic process is also isentropic (constant  entropy).    Marand’s Notes: Chapter 3 ‐ The Second Law of Thermodynamics  99  3)  Reversible Heating at Constant Pressure (all materials)  dS = δq / T = δqP / T = dH / T = CP dT / T  Therefore, ΔS = CP ln(Tf / Ti)    4)  Reversible Heating at Constant Volume (all materials)  dS = δq / T = δqV / T = dU / T = CV dT / T  Therefore, ΔS = CV ln(Tf / Ti)    5)  Reversible Phase Transformations:  If a phase transformation takes place under equilibrium conditions, then   dS = δq /T = dH / T  since during a phase transformation, the temperature (as well as the  pressure, see above δq = dH) are constant. Then we can integrate dS to  obtain:  ΔS = ΔH...
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## This note was uploaded on 01/26/2014 for the course CHEM 3615 taught by Professor Aresker during the Spring '07 term at Virginia Tech.

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