5.60 Spring 2007 • Lecture #5 page 1 Reversible Adiabatic Expansion (or compression) of an Ideal Gas 1 mole gas (V1,T1) = 1 mole gas (V2,T2) đq = 0 Ideal gas ⇒ adiabatic ⇒ From 1st Law ∴ CV dT = − pdV T2 CV ∫ T1 ⇒ p =RT V V dV dT = −R ∫ V V T 2 dU = -pd V ⇒ Define CvdT = -pdV along path dT dV = −R T V CV ⇒ 1 Cp γ ≡ CV đ w = -pd V Reversible ⇒ d U = Cv d T ⇒ ⎛T2 ⎞ ⎛V1 ⎞ ⎜ ⎟=⎜ ⎟ ⎝T1 ⎠ ⎝V2 ⎠ Cp R CV ⎛T2 ⎞ ⎛V1 ⎞ ⎜ ⎟=⎜ ⎟ ⎝T1 ⎠ ⎝V2 ⎠ C p −CV =R for i.g. ⎯⎯⎯⎯⎯⎯→ ⎛T2 ⎞ ⎛V1 ⎞CV ⎜ ⎟=⎜ ⎟ ⎝T1 ⎠ ⎝V2 ⎠ −1 γ −1 3 ⎫ For monatomic ideal gas: CV = R ⎪ 2 ⎪ 5 ( > 1 generally) ⎬ γ = 5 ⎪ 3 Cp = R 2 ⎪⎭ In an adiabatic expansion (V2 > V1), the gas cools (T2 > T1). And in an adiabatic compression (V2 < V1), the gas heats up. For an ideal gas (one mole) pV γ pV T = R γ ⇒ ⎛ p2 ⎞ ⎛V1 ⎞ ⎜ ⎟=⎜ ⎟ ⎝ p1 ⎠ ⎝V2 ⎠ γ γ ⇒ pV 1 1 = pV 2 2 is constant along a reversible adiabat For an isothermal process T = constant ⇒ pV = constant 5.60 Spring 2007 Lecture #5 page 2 p V2adiabat < V2isotherm because the gas cools during reversible adiabatic expansion p1 p2 V2ad V2iso V1 • Irreversible Adiabatic Expansion of an ideal gas against a constant external pressure 1 mol gas (p1,T1) = 1 mol gas (p2,T2) adiabatic Constant pext = p2 Ideal gas 1st Law ∴ Integrating: ⇒ ⇒ ⇒ ⇒ đq = 0 đw = -p2d V d U = Cv d T dU = -p2 dV Cv d T = - p 2 dV Cv ( T2 - T 1 ) = - p 2 ( V2 - V1 ) ⎛ Using pV = RT Note (pext=p2) p2 < p1 T2 (CV + R ) =T1 ⎜ CV + ⎝ ⇒ Note also (-wrev) > (-wirrev) T2 < T1 p2 ⎞ R⎟ p1 ⎠ Again, expansion cools Less work is recovered through an irreversible process 5.60 Spring 2007 Lecture #5 page 3 Some Thermodynamic Cycles • Reversible Ideal Gas processes: Find ΔU , ΔH , q , w , p p1 p2 p3 (I) (T1) B V1 [A] (T1) C (const. V) V wA = −RT1 ln 2 V1 1 mol gas (p1,V1,T1) Ideal gas: 1st Law: (T3) E (const. V) A p2 (isotherm) V2 1 mol gas (p1,V1,T1) Adiabat: D (const. p) (T1) (T2) Ideal gas isotherm: [B] (T1) p1 A (isotherm) (rev. adiabat) (II) p V1 const. T = 1 mol gas (p2,V2,T1) ΔU A = 0 V qA = RT1 ln 2 V1 rev.adiabat = V2 ΔH A = 0 đq ∫T 1 mol gas (p3,V3,T2) qB = 0 ΔUB = CV (T2 −T1 ) ΔHB = C p (T2 −T1 ) wB = CV (T2 −T1 ) V2 V1 =R ln đqB ∫T =0 đq ∫T 5.60 Spring 2007 Lecture #5 1 mol gas (p3,V2,T2) [C] reversible wC = 0 1st Law: qC = CV (T1 −T2 ) đqC ∫T vs. ΔUA = 0 ΔH A = 0 ΔUC = CV (T1 −T2 ) ΔHC = C p (T1 −T2 ) [B] + [C] ΔUB + ΔUC = 0 = ΔUA ΔHB + ΔH C = 0 = ΔH A V qA = RT1 ln 2 V1 V w A = −RT1 ln 2 V1 đqA Ideal gas: ⎛T ⎞ = CV ln ⎜ 1 ⎟ ⎝T2 ⎠ [A] ∫T 1 mol gas (p2,V2,T1) = const. V Constant V: page 4 V2 V1 = R ln qB + qC = CV (T1 −T2 ) ≠ qA w B + w C = CV (T2 −T1 ) ≠ w A đqB ∫T +∫ đqC T ⎛V ⎞ đq = R ln ⎜ 2 ⎟ = ∫ A T ⎝V1 ⎠ ⎛ đq ⎞ This result suggests that ⎜ ∫ rev ⎟ is a state function! ⎝ T ⎠ [D] ΔUD = CV (T3 −T1 ) ΔHD = C p (T3 −T1 ) wD = −R (T3 −T1 ) qD = C p (T3 −T1 ) đqD ∫T ⎛T ⎞ = C p ln ⎜ 3 ⎟ ⎝T1 ⎠ 5.60 Spring 2007 Lecture #5 ΔUE = CV (T1 −T3 ) ΔHE = C p (T1 −T3 ) [E] wE = 0 đqE ∫T [A] [D] + [E] ΔUD + ΔUE = ΔUA ΔHD + ΔHE = ΔHA V qA = RT1 ln 2 V1 V w A = −RT1 ln 2 V1 đqA ∫T qE = CV (T1 −T3 ) ⎛T ⎞ = CV ln ⎜ 1 ⎟ ⎝T3 ⎠ vs. ΔUA = 0 ΔH A = 0 page 5 V2 V1 = R ln qD + qE = R (T3 −T1 ) ≠ qA wD + w E = −R (T3 −T1 ) ≠ w A đqD ∫T +∫ đqE T ⎛V ⎞ đq = R ln ⎜ 2 ⎟ = ∫ A T ⎝V1 ⎠ ⎛ đq ⎞ Here again ⎜ ∫ rev ⎟ looks like a state function! ⎝ T ⎠