general chemistry4

60 spring 2004 lecture 6 page 4 thermodynamic cycles

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Unformatted text preview: eversible Ideal Gas processes: Calculate p p1 p2 p3 () I () T 1 p p1 (const. ) V () II () T 1 B A (isotherm) () T 1 () T 2 D (const.p) A (isotherm) () T 3 (rev. adiabat) C p2 E (const.V) () T 1 V1 V2 V1 V2 = [A] 1 mol gas (p1,V1,T1) Ideal gas isotherm: 1ST Law: Reversible: Ideal gas: wA = - p dV = -RT1 V1 V2 V2 V1 T const. 1 mol gas (p2,V2,T1) DUA = 0 dq = - dw dw = - pdV dV V = - RT1 ln 2 = -q A V V1 DH A = 0 wA = -RT1 ln qA = RT1 ln V2 V1 qA = RT1 V2 V1 dV V = RT1 ln 2 V V1 2 2 V2 V1 dqA T = 1 V V dV ^ V dV RT1 V = R V = R ln 2 T1 V V V1 1 1 dqA T = Rln V2 V1 [B] 1 mol gas (p1,V1,T1) qB = 0 rev.adiabat = T2 1 1 mol gas (p3,V3,T2) Adiabat: Ideal gas: 1st Law: DU B = T CV dT = CV (T2 - T1 ) DH B = T C p dT = C p (T2 - T1 ) 1 DUB = CV (T2 - T1 ) DHB = C p (T2 - T1 ) T2 wB = CV (T2 - T1 ) dqB =0 T 5.60 Spring 2004 Lecture #6 page 5 [C] 1 mol gas (p3,V2,T2) wC = 0 reversible const. V = 1 mol gas (p2,V2,T1) Constant V: 1st Law: Ideal g...
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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.

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