Ideales Gasgesetz, Isentropenexponent Luft

118 days ago by Mebus

from scipy.constants import R, N_A from sage.misc.latex import pretty_print # Einheiten mol = units.amount_of_substance.mole Pa = units.pressure.pascal m = (units.length.meter) K = units.temperature.kelvin g = units.mass.gram J = units.energy.joule kg = g*10^3 R = R*J/(K*mol) ML = 28.964*g/mol cpWig = 1.86*J/(mol*K) cpWl = 4,18*J/(mol*K) cpLig = 1*J/(mol*K) # --------------------------------------------------------------------- # Parameter p = 10^5*Pa T = 293.15*K masse = 1*kg n = masse/ML print "\n1 kg Luft bei Standardbedingungen:" V = n*R*T/p pretty_print (V.convert()) print "\n1 mol Luft:" n = 1*mol V = n*R*T/p pretty_print (V.convert()) # ------------------------------------------------------------------------ igg = p*V-n*R*T V=solve(igg,V) #print V[0].convert() print "\nEnthalpie Wasser bei 20 °C (iF):" print (cpWig*293.15*K) print "\nIsentropenexponent von Luft:" cvLig = R-cpLig print (cpLig/cvLig) 
       
1 kg Luft bei Standardbedingungen:
\newcommand{\Bold}[1]{\mathbf{#1}}0.841523086175943 \, \mbox{meter}^{3}

1 mol Luft:
\newcommand{\Bold}[1]{\mathbf{#1}}0.0243738746680000 \, \mbox{meter}^{3}

Enthalpie Wasser bei 20 °C (iF):
545.259000000000*joule/mole

Isentropenexponent von Luft:
0.13671526803301728
1 kg Luft bei Standardbedingungen:
\newcommand{\Bold}[1]{\mathbf{#1}}0.841523086175943 \, \mbox{meter}^{3}

1 mol Luft:
\newcommand{\Bold}[1]{\mathbf{#1}}0.0243738746680000 \, \mbox{meter}^{3}

Enthalpie Wasser bei 20 °C (iF):
545.259000000000*joule/mole

Isentropenexponent von Luft:
0.13671526803301728