First Law and Ideal Gases - Colby College

145 downloads 225 Views 24KB Size Report
Colby College. Application of the First Law to Ideal Gases. Calculate q,w, ∆U, and ∆H for ideal gas processes: dU = ñq + ñw. dU = Cv dT dH = Cp dT. ∆U = q + w.
Application of the First Law to Ideal Gases Calculate q,w, ∆U, and ∆H for ideal gas processes: dU = ñq + ñw dU = Cv dT dH = Cp dT ∂U ∆U = q + w for any process since   = 0 ∂VT Isothermal Reversible Expansion dT = 0 so dU = and dH = q = –w V2  V1

w = – nRT ln  P2V2 = P1 V1

V2 P1 V1 = P2

or

P1 w = – nRT lnP  

2

Adiabatic Reversible Expansion ñq = so dU = ñw = – P dV dU = – P dV or dU = Cv dT Cv dT = – P dV dT dV Cv T = – nR V dT dV ⌠T 2 ⌠V = –  2 nR  Cv T V ⌡T 1 ⌡ V1 V2

T2

Cv ln T

|T

= – nR ln V 1

|V

1

T2 V2 Cv ln  = – nR ln  T1 V1

Cv T2 V2 Method 1: nR lnT  = – lnV  c = Cv/nR  1  1 Cv/nR Cv/nR T2Cv/nR V1     or = V = V 2T2 1T1 T1 V2

Colby College

First Law and Ideal Gases

Method 2: Reversible adiabatic: P = Pext dH = dU + PdV + VdP = ñq – P dV + PdV + VdP = V dP dH = CpdT = V dP

dP CpdT = nRT P

T2 P2 Cp ln T  = nR lnP 

Cp T2 P2   = ln  ln nR T1 P1



1



1

T2Cp/nR P2   =  T1 P1

Relate cst V and cst P processes: Cp ln(T2/T1) nR ln (P2/P1) γ = Cp/Cv = Cv ln(T2/T1) – nR ln(V2/V1) γ=

ln (P2/P1) ln(V1/V2)

γ ln(V1/V2) = ln(P2/P1) V1γ P2   =  V2 P1 γ

P2V2 = P1Vγ1

P1 8

P (bar) 6 isothermal

P2V2 = P1V1

4 2

adiabatic γ

P2V2 = P1V1

0

V1

2

V (L)

γ 4

Adiabatic Irreversible Expansion: Example: Pext = constant = P2 ∆U = w ∆U = Cv (T2 – T1) ∆U = – Pext (V2 – V1) Cv ∆T = – Pext ∆V Cv(T2 – T1) = – Pext(V2 – V1) nRT2 nRT1 Cv(T2 – T1) = – P2  P – P   2 1  monatomic gas: Cv=

Colby College

3 nR 2

5 diatomic gas (neglecting vibration): Cv= nR 2

V2

6