University:
Massachusetts Institute of Technology
Course:
20.110J | Thermodynamics of Biomolecular Systems
Academic year:

2022
Views:

353

Pages:

5
Author:

Efrain Villa

Lecture #4 page 1 EXPANSIONS, THERMODYNAMIC CYCLES • 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 Define 2 dU = -pd V ⇒ CV đw = -pd V CvdT = -pdV along path dT dV = −R T V R CV ⇒ 1 Cp γ ≡ CV Reversible ⇒ dU = Cv d T ⎛T2 ⎞ ⎛V1 ⎞ ⎜ ⎟=⎜ ⎟ ⎝T1 ⎠ ⎝V2 ⎠ Cp C p −CV =R for i.g. ⎯⎯⎯⎯⎯⎯→ ⎛T2 ⎞ ⎛V1 ⎞CV ⎜ ⎟=⎜ ⎟ ⎝T1 ⎠ ⎝V2 ⎠ γ −1 ⇒ ⎛T2 ⎞ ⎛V1 ⎞ ⎜ ⎟=⎜ ⎟ ⎝T1 ⎠ ⎝V2 ⎠ 3 ⎫ For monatomic ideal gas: CV = R ⎪ 2 ⎪ 5 ( > 1 generally) ⎬ γ = 5 ⎪ 3 Cp = R 2 ⎪⎭ [More generally the heat capacity for a molecular species has a temperature dependence that can be approximated as C p (T ) = a + bT + cT 2 with a, b, and c tabulated.] In an adiabatic expansion (V2 > V1), the gas cools (T2 > T1). And in an adiabatic compression (V2 < V1), the gas heats up. −1 Lecture #4 pV T = R For an ideal gas (one mole) pV γ page 2 γ ⇒ ⎛ p2 ⎞ ⎛V1 ⎞ ⎜ ⎟=⎜ ⎟ ⎝ p1 ⎠ ⎝V2 ⎠ γ γ ⇒ pV 1 1 = pV 2 2 is constant along a reversible adiabat Recall, for an isothermal process T = constant ⇒ pV = constant p isotherm pV=constant p1 pVγ=const. p2 adiabat V2adiabat < V2isotherm because the gas cools during reversible adiabatic expansion 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: Using pV = RT ⇒ ⇒ ⇒ ⇒ (pext=p2) đq = 0 đw = -p2d V dU = Cv d T dU = -p2 dV Cv d T = - p 2 dV Cv ( T2 - T 1 ) = - p 2 ( V2 - V1 ) ⎛ T2 (CV + R ) =T1 ⎜ CV + ⎝ p2 ⎞ R⎟ p1 ⎠ Lecture #4 p2 < p1 Note ⇒ T2 < T1 (-wrev) > (-wirrev) Note also page 3 Again, expansion cools as expected, less work is recovered through an irreversible process Thermodynamic Cycles • Reversible Ideal Gas processes: Find ∆U , ∆H , q , w p p1 (I) (T1) p1 A (isotherm) p2 p3 B (T1) (rev. adiabat) (T2) V1 (II) p C (const. V) p2 V2 (T1) D (const. p ) A (isotherm) V1 (T3) E (const. V) (T1) V2 For (I) 1gas(p1,V1,T1) A 1gas(p2,V2,T1) B C 1gas(p3,V2,T2) There are two paths from initial to final states (A) and (B+C). As far as functions of states (e.g. U, H) are concerned it doesn’t matter which path is taken. Lecture #4 [A] 1 mol gas (p1,V1,T1) const. T = V wA = −RT1 ln 2 V1 [B] 1 mol gas (p1,V1,T1) Adiabat: Ideal gas: 1st Law: [C] 1 mol gas (p2,V2,T1) ∆UA = 0 Ideal gas isotherm: page 4 ∆HA = 0 V qA = RT1 ln 2 V1 rev.adiabat = 1 mol gas (p3,V3,T2) qB = 0 ∆UB = CV (T2 −T1 ) ∆HB = C p (T2 −T1 ) w B = CV (T2 −T1 ) 1 mol gas (p3,V2,T2) reversible = const. V 1 mol gas (p2,V2,T1) Constant V: wC = 0 1st Law: qC = CV (T1 −T2 ) Ideal gas: ∆UC = CV (T1 −T2 ) ∆HC = C p (T1 −T2 ) Lecture #4 [A] vs. ∆UA = 0 ∆HA = 0 qB + qC = CV (T1 −T2 ) ≠ qA wB + w C = CV (T2 −T1 ) ≠ w A ∆UD = CV (T3 −T1 ) ∆HD = C p (T3 −T1 ) ∆UE = CV (T1 −T3 ) ∆HE = C p (T1 −T3 ) [E] [A] ∆UA = 0 ∆HA = 0 V qA = RT1 ln 2 V1 V w A = −RT1 ln 2 V1 [B] + [C] ∆UB + ∆UC = 0 = ∆UA ∆HB + ∆HC = 0 = ∆HA V qA = RT1 ln 2 V1 V w A = −RT1 ln 2 V1 [D] page 5 vs. qD = C p (T3 −T1 ) wE = 0 wD = −R (T3 −T1 ) qE = CV (T1 −T3 ) [D] + [E] ∆UD + ∆UE = ∆UA ∆HD + ∆HE = ∆HA qD + qE = R (T3 −T1 ) ≠ qA wD + w E = −R (T3 −T1 ) ≠ w A

Reversible and Irreversible Processes

Related Documents

Recommended Documents

Description

Related Documents

Free up your schedule!

Take 5 seconds to unlock