a) folosim formula niu=m/μ=N/NA=V/Vμ
de aici extragem m/μ=N/NA => m=(μ*NA)/N=(2*28*10⁻³*6,022*10²³)/60,2*10²³≈56*10⁻⁴ kg
b)ρ=m/V
extragem din formula N/NA=V/Vμ => V=(N*Vμ)/NA=(60,2*10²³*22,4*10⁻³)/6,022*10²³=223,92*10⁻³ m³
ρ=m/V=(56*10⁻⁴)/223,92*10⁻³=0,25*10⁻¹=25*10⁻³ kg/m³
c)n=N/V=(60,2*10²³)/223,92*10⁻³=0,26*10²⁶ molecule/m³
d) p'V=niu'*R*T
daca nr de molecule se schimba,atunci se schimba automat si nr de moli,masa,si volumul gazului
scriu din nou formula lui niu si extrag
niu'=N'/NA=20%N/NA=(0,2*60,2*10²³)/6,022*10²³=1,99≈2 moli
p'V'=niu'R*T
aflu si volumul final
niu'=V'/Vμ => V'=niu'*Vμ=2*22,4*10⁻³=44,8*10⁻³ m³
p'V'=niu'RT
p'=(niu'RT)/V'=(2*8,31*350)/44,8*10⁻³=129,84*10³ Pa
