Răspuns:
V=40L
m=140g \
miu,N2=28g/mol
niu=m/miu=140/28mol= 5mol
T=27+273=300K
pV=niuRT--> pX40=5X0,082X300---> p=3,075atm
3,69X40=(5+ niu,CO2)X0,082X300---> (5+niu,CO2)=6-->niu,CO2=1mol
miu,mediu= x.N2Xmiu,N2+ x.CO2Xmiu,CO2 x-FRACTII MOLARE
x,N2=5/6; miu,N2=28g/mol
x,CO2= 1/6; miu,CO2=44g/mol
miu, mediu= 5/6X28+1/6X44=>30.66g/mol
d=miu,mediu/miu,aer=30,66/29=1,06
Explicație:
