Question 12.179E: A 4 lbm mixture of 50% argon and 50% nitrogen by mole is in ...

A 4 lbm mixture of 50% argon and 50% nitrogen by mole is in a tank at 300 psia, 320 R. How large is the volume using a model of (a) ideal gas and (b) Kays rule with generalized compressibility charts.

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a) Ideal gas mixture
Eq.11.5:

\begin{aligned}& M _{\operatorname{mix}}=\sum y _{ i } M _{ i }=0.5 \times 39.948+0.5 \times 28.013=33.981 \\& R =1545.36 / 33.981=45.477   lbf – ft / lbm – R \\& V =\frac{ mRT }{ P }=\frac{4 \times 45.477 \times 320}{300 \times 144} \frac{ lbf – ft }{ psi \times in ^{2} / ft ^{2}}= 1 . 3 4 7   \mathrm { ft } ^ { 3 }\end{aligned}

 

b) Kay’s rule Eq.12.84

\begin{aligned}& P _{ c  \text { mix } }=0.5 \times 706+0.5 \times 492=599   psia \\& T _{ c  \text { mix }}=0.5 \times 271.4+0.5 \times 227.2=249.3   R\end{aligned}

Reduced properties:      P _{ r }=\frac{300}{599}=0.50, \quad T _{ r }=\frac{320}{249.3}=1.284

Fig. D.1:          Z = 0.92

V = Z \frac{ m \overline{ R } T }{ M _{\operatorname{mix}} P }= Z V _{ ID  gas }=0.92 \times 1.347= 1 . 2 4   ft ^{3}

………………………………………….

Eq.11.5: M_{\operatorname{mix}}=\frac{m_{ tot }}{n_{ tot }}=\frac{\sum n_{i} M_{i}}{n_{ tot }}=\sum y_{i} M_{i}

Eq.12.84 : \left(P_{c}\right)_{\operatorname{mix}}=\sum_{i} y_{i} P_{c i}, \quad\left(T_{c}\right)_{\operatorname{mix}}=\sum_{i} y_{i} T_{c i}

 

 

D.1

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