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Blog (2007-05-29)
Mathe,Physik,Chemie

Mathe,Physik,Chemie

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19892031 Antworten ...
 
Eine simple Frage











18015325 Antworten ...
 
an alle










20 Grad Celsius = 273.15 + 20





p*V= n*R*T






8.314 J / (mol * Kelvin)




1 mol CO2 = 12 + 2 * 18 = 48 gr

n = 1000 gr / 48 gr = 20.83 mol CO2


V = n * R * T / p

V =

(20.83 mol )* {8.314 J / (mol * Kelvin)} * 293.15 Kelvin / 101325 Pa


V =
0.501 J / Pa
V =
0.501 N * m / ( N / m2)
V =
0.501 m3 = 501 Liter CO2 als Gas


.
17997130 Antworten ...
 
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