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Question 11.184E: Weighing of masses gives a mixture at 80 F, 35 lbf/in.^2 wit...

Weighing of masses gives a mixture at 80 F, 35 lbf / in .^{2} with 1 lbm O _{2}, 3 lbm N _{2} and 1 lbm CH _{4}. Find the partial pressures of each component, the mixture specific volume (mass basis), mixture molecular weight and the total volume.

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From Eq. 11.4:          y _{ i }=\left( m _{ i } / M _{ i }\right) / \sum m _{ j } / M _{ j }

n _{ tot }=\sum m _{ j } / M _{ j }=(1 / 31.999)+(3 / 28.013)+(1 / 16.04)

= 0.031251 + 0.107093 + 0.062344 = 0.200688

\begin{aligned}& y _{ O 2}=0.031251 / 0.200688=0.1557, \quad y _{ N 2}=0.107093 / 0.200688=0.5336, \\& y _{ CH 4}=0.062344 / 0.200688=0.3107\end{aligned}

From Eq.11.10:

\begin{aligned}&P_{O 2}=y_{O 2} P_{t o t}=0.1557 \times 35=5.45   lbf / \text { in. }^{2} \\&P_{N 2}=y_{N 2} P_{t o t}=0.5336 \times 35=18.676   lbf / \text { in. }^{2} \\&P_{C H 4}=y_{C H 4} P_{t o t}=0.3107 \times 35=10.875   lbf / \text { in. }^{2}\end{aligned} \begin{aligned}& V _{\text {tot }}= n _{\text {tot }} \overline{ R } T / P =0.200688 \times 1545 \times 539.7 /(35 \times 144)= 3 3 . 2   ft ^{3} \\& v = V _{\text {tot }} / m _{\text {tot }}=33.2 /(1+3+1)= 6 . 6 4 ~ f t ^{ 3 } / lbm\end{aligned}

From Eq.11.5:

M _{\operatorname{mix}}=\sum y _{j} M _{ j }= m _{ tot } / n _{ tot }=5 / 0.200688= 2 4 . 9 1 4   lbm / lbmol

…………………………………………

Eq.11.4 : y_{i}=\frac{n_{i}}{n_{ tot }}=\frac{m_{i} / M_{i}}{\sum m_{j} / M_{j}}=\frac{m_{i} /\left(M_{i} m_{ tot }\right)}{\sum m_{j} /\left(M_{j} m_{ tot }\right)}=\frac{c_{i} / M_{i}}{\sum c_{j} / M_{j}}

Eq.11.10 :  P_{A}=y_{A}  P, \quad P_{B}=y_{B}  P

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

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