Question 14.11: Ionization and Neutral Component A mixture containing 95, he...

Helium ionization can be completely ignored, given the high value of the equilibrium constant. Set up the overall reaction and use the equilibrium equation for the ionization of cesium to obtain the desired pressure.

Assumptions
The IG mixture model is applicable. The IG model is applicable to electrons. Ionization of cesium is the only elementary step.

Analysis
The overall reaction can be expressed in terms of 1 kmol of helium:

He +(5 / 95) Cs \rightarrow He +a Cs +b Cs ^{+}+c e ^{-}

For each Cs ion, there must be one free electron; therefore, b = c. Also, using atom balance equation for Cs, the overall reaction can be expressed in terms of one unknown:

He +0.053 Cs \rightarrow He +a Cs +(0.053-a) Cs ^{+}+(0.053-a) e ^{-}

The mole fraction of electrons in the mixture is

\begin{aligned}&y_{ e ^{-}}=\frac{0.053-a}{1.105-a}=0.01 ; \quad \Rightarrow \quad a=0.0411 \\&\Rightarrow \quad y_{ Cs }=\frac{a}{1.105-a}=0.0386 ; \quad \text { and } \quad y_{ Cs ^{+}}=y_{ e ^{-}}=0.01\end{aligned}

Substituting these expressions in the equilibrium equation, Eq. (14.50), for the ionization reaction Cs \rightarrow Cs ^{+}+ e ^{-} :

K \equiv \frac{\prod\limits_{p} y_{p}^{\nu_{p}}}{\prod\limits_{r} y_{r}^{\nu_{r}}}\left(\frac{p}{p_{0}}\right)^{\sum\limits_{p} v_{p}-\sum\limits_{r} v_{r}}                   (14.50)

\begin{aligned}&K=\frac{y_{ Cs ^{+} y_{ e }^{-}}}{y_{ Cs }}\left(\frac{p}{p_{0}}\right)^{1+1-1}=\frac{y_{ e ^{-}}^{2}}{y_{ Cs }} \frac{p}{p_{0}}=\frac{0.01^{2}}{0.0386} \frac{p}{p_{0}} ;\\&\Rightarrow \quad p=\frac{p_{0} K}{0.00259}=\frac{(100)(15.63)}{(0.00259)}=603.5 MPa\end{aligned}

TEST Analysis
The IGE model currently does not have Cs ^{+} in its database; therefore, a TEST Solution is not possible at this time.

Discussion
The MHD (magneto hydrodynamics) converters takes advantage of this type of ionization to produce electricity by passing the ionized gases through a magnetic field (Fig. 14.32).