zum Directory-modus

Lösung

V = 0,5L = 0,5dm3 = 5⋅10−4 m3 T = K °C θ + 273 K = 373 K M Benzol = 78 g mol p = n R T V = m R T M V = 10 g ⋅ 8,3 J mol ⋅ K ⋅ 373 K 78 g mol ⋅ 5⋅10−4 m3 = 10 ⋅ 8,3 ⋅ 373 78 ⋅ 5 ⋅ 104 ⋅ g ⋅ J mol ⋅ K ⋅ K g mol ⋅ m3 = 793.800 J m3 = 793.800 Pa 1 J m3 = 1 N m m3 = 1 N m2 = 1Pa