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Legile gazelor
V=98,4 m cubi de amestec gazos ( 02,CO2,N2)
t=127°C
p=2 atmosfere
Daca avem 6,022×10^26 molecule de CO2 si 64 kg O2 restul fiind N
Sa se calculeze: 1. cantitatea de N din recipient
2. masa molara medie
3. masa gazelor din recipient

Răspuns :

Puia68
V= 98,4x10³ l;  p=2 atm;   T= 127+273=400 grd Kelvin
pV= niu RT---> niu=2x98,4x10³/ 0,082x400-----> niu=6000 moli gaze

niu,CO2= N/N,Avogadro;
niu=6,023x10²⁶molec/ 6,023x10²³molec/mol=1000molCO2
niu,O2= m/M
niu= 64kg/32kg/kmol= 2kmol=2000mol O2
niu,N2= 6000-3000= >3000molN2
2.
M,mediu= 1000/6000xM,CO2+2000/6000xM,O2+ 3000/6000M,N2
M=44gx1/6 +32gx2/6 + 28gx3/6   CALCULEAZA  !!!

3. m= 1000molx44g/mol + 2000molx 32g/mol + 3000molx24g/mol
   CALCULEAZA !!!!!

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