Question #213406

The pressure and temperature of mixture of 4 kg of O2 and 6 kg of N2 are 4 bar 

and 27°C respectively. For the mixture determine the following ) :i) The mole 

fraction of each component ; (ii) The average molecular weight; (iii) The specific 

gas constant ; (iv) The volume and density ;(v) The partial pressures and partial 

volumes


1
Expert's answer
2021-07-12T02:12:02-0400

Mole fraction of oxygen

no2=gmweightmoleculerweight=432=0.125n_{o_2}=\frac{gm weight}{moleculer weight}=\frac{4}{32}=0.125

nN2=gmweightmoleculerweight=628=0.214n_{N_2}=\frac{gm weight}{moleculer weight}=\frac{6}{28}=0.214

Molecular weight

Molecular weight for

oxygen =2×\times atomic weights

(Wm)o2=2×16=32(W_m)_{o_2}=2\times16=32

Nitrogen molecular weight

=2×2\times atomic weights

(Wm)N2=2×14=28(W_m)_{N_2}=2\times14=28

Specific gas constan

Rs=RMo2=8.3132=0.2596R_s=\frac{R}{M{_o}_{2}}=\frac{8.31}{32}=0.2596

Rs=RMN2=8.3128=0.296R_s=\frac{R}{M{_N}_{2}}=\frac{8.31}{28}=0.296

Volume and density

V=nRTPV=\frac{nRT}{P}

Vo=0.125×8.314×3004=77.94m3V_o=\frac{0.125\times8.314\times300}{4}=77.94m^3

VN=0.214×8.314×3004=133.43m3V_N=\frac{0.214\times8.314\times300}{4}=133.43m^3

do=MV=3277.94=0.4105kg/m3d_o=\frac{M}{V}=\frac{32}{77.94}=0.4105kg/m^3

do=MV=28133.43=0.2098kg/m3d_o=\frac{M}{V}=\frac{28}{133.43}=0.2098kg/m^3

Partial pressure

Po=n×P_o=n\times pressure

Po=0.125×4=0.5barP_o=0.125\times4=0.5 bar

PN=n×P_N=n\times pressure

PN=P_N= 0.214×4=0.8560.214\times4=0.856 bar

Partial volume

Vo=n×V_o=n\times volume

Vo=0.125×77.94=9.74kgV_o=0.125\times77.94=9.74kg

VN=n×V_N=n\times volume

VN=0.214×133.43=28.55kgV_N=0.214\times133.43=28.55kg


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