Lösung von Aufgabe 15
- thermische Größen:
- p1 = 1 bar, V1 = 0.01
m3, T1 = 288 K
- V2 = 0.001 m3
- p2 = p1
(V1/V2)κ = 25.12 bar
- T2 = T1
(V1/V2)(κ - 1) = 723.8 K
- V3 = 10-4 m3,
p3 = 631.0 bar, T3 = 1818 K
- Volumenänderungsarbeit:
- WV12 = (p1
V1)/(κ - 1) [
(V1/V2)(κ - 1) - 1 ]
- = 3.780 kJ
- WV23 = 9.494 kJ
- kalorische Größen:
- U2 - U1 = WV12 =
3.780 kJ
- U3 - U2 = WV23 =
9.494 kJ
- H2 - H1 = κ
(U2 - U1) = 5.292 kJ
- H3 - H2 = 13.29 kJ