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Unformatted text preview: heats up. For ideal gas (one mole)
pV T= R p ^ V ^ fi 2~ = 1~ p1 V2 g In other words p1V1g = p2V2g pV g is constant along a reversible adiabat
For isothermal process T = constantfi pV = constant pV is constant along an isotherm p
p1 p2 V1 V2ad V2iso V2adiabat < V2isotherm because the gas cools during reversible adiabatic expansion 5.60 Spring 2004 Lecture #6 page 3 Irreversible Adiabatic Expansion of an ideal gas against constant external pressure one mole ideal gas irreversible adiabatic constant pext 1 mol gas (p1,T1) = 1 mol gas (p2,T2) Adiabatic Constant pext = p2 Ideal gas 1st Law \ Integrating: Using pV = RT p ^ T2 (CV + R ) = T1 CV + 2 R~ p1 fi fi fi fi dq = 0 dw = - pdV
dU = CvdT dU = -p2dV CvdT = - p2dV Cv (T2 - T1) = - p2 (V2 - V1) Note p2 < p1 fiT2 < T1 Again, an expansion cools the gas Note also (-wrev) > (-wirrev) Less work recovered through the irreversible process 5.60 Spring 2004 Lecture #6 page 4 Thermodynamic Cycles
dq DU , DH , q, w, T R...
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This note was uploaded on 04/23/2009 for the course GENERAL CH 1230 taught by Professor Unknown during the Spring '04 term at Toledo.
- Spring '04