Examen

Problema 1.-
a) Nombre de cada proceso
(1-2) → Expansión isentropica
(2-3) → Enfriamiento isocórico
(3-4) → Compresión politrópica
(4-1) → Calentamiento Isocórico
b) Valores PVTPunto | P (Bar) | V(m3) | T(K) | n (mol) |
1 | 300 | 0.04157 | 696.44 | 0.21538 |
2 | 43.076 | 0.16628 | 400 | 0.21538 |
3 | 21.538 | 0.16628 | 200 | 0.21538 |
4 | 200 | 0.04157 | 464.296 |0.21538 |

1-2 expansión adiabática
P2=P1V1V2γ= 300(0.041570.16628)1.4 = 43.076 Bar
n = P2 V2RT2= 43.076*0.166280.08314*400 = 0.21538 mol
T1 = P1 V1Rn= 300*0.041570.0814* 0.21538=696.44 KP3=T3T2P2 =200400*43.076=21.538 Bar
T4 = P3 V3Rn= 200*0.041570.0814* 0.21538=464.296 K
c) Exponente politrópico
δ 3-4= lnP4P3lnV3V4= ln20021.538ln0.166280.04157=1.6075

d) Valores de W,Q, ∆U, ∆Hy ∆S.
Gas diatomico cv = 20.785 KJ/mol K
Cp = 29.099 KJ/mol K
(1-2) expansión isentrópica
W = ∆U = nCv∆T = (0.21538 * 20.785)(400-696.44) = -1327.065 KJ
Q = ∆S =0
∆H = ∆U*γ = -1327.065* 1.4 = -1857.891 KJ
(2-3) Enf. Isocórico
Q = ∆U = nCv∆T = (0.21538 * 20.785)(200-400) = -895.335 KJ
∆H = ∆U*γ = -895.335* 1.4 = 1253.468 KJ
∆S = nCvlnT3T2=0.21538*20.785*ln200400= -3.1030KJK
W = 0
(3-4) compresión politrópica
∆U = nCv∆T = (0.21538 * 20.785)(464.296-200) = 1183.167 KJ
∆H = ∆U* 1.4 = 1656.434 KJ
∆S = nCvlnT4T3= (0.21538 * 20.785)ln464.296200=1.2879KJK
Q = ∆U – W = 1183.167 - 779.040 = 404.127 KJ
W = nRδ-1(∆T) = 0.21538(8.413)1.6075-1(464.296-200) = 779.040 KJ
(4-1) calentamiento isocorico
Q = ∆U = nCv∆T = (0.21538 *20.785)(696.44-464.296) = 1039.233 KJ
∆H = ∆U*γ = 1039.233 * 1.4 = 1454.926 KJ
∆S = nCvlnT1T4= 0.21538*20.785*ln696.44464.296= 1.8151KJK
W = 0

Proceso | ∆U(KJ) | ∆H(KJ) | ∆S(KJ/K) | W(KJ) | Q(KJ) |
1-2| -1327.065 | -1857.891 | 0 | -1327.065 | 0 |
2-3 | -895.335 | -1253.468 | -3.1030 | 0 | -895.335 |
3-4 | 1183.167 | 1650.434 | 1.2879 | 779.040 | 404.127 |
4-1 | 1039.233 | 1454.926 | 1.8151...