Ionization and Neutral Component A mixture containing 95, helium and 5, cesium on molar basis is heated to 2000 K. The equilibrium constants for ionization of helium and cesium are 5000 and 15.63, respectively. Determine the pressure at which the mole fraction of electron’s is 0.01.

Question

Ionization and Neutral Component

A mixture containing 95% helium and 5% cesium on molar basis is heated to 2000 K. The equilibrium constants for ionization of helium and cesium are 5000 and 15.63, respectively. Determine the pressure at which the mole fraction of electron’s is 0.01.

Accepted Answer

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:

[latex] He +(5 / 95) Cs ightarrow He +a Cs +b Cs ^{+}+c e ^{-} [/latex]

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:

[latex] He +0.053 Cs ightarrow He +a Cs +(0.053-a) Cs ^{+}+(0.053-a) e ^{-} [/latex]

The mole fraction of electrons in the mixture is

[latex]\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 ext { and } \quad y_{ Cs ^{+}}=y_{ e ^{-}}=0.01\end{aligned}[/latex]

Substituting these expressions in the equilibrium equation, Eq. (14.50), for the ionization reaction [latex] Cs ightarrow Cs ^{+}+ e ^{-} [/latex]:

[latex] K \equiv \frac{\prod\limits_{p} y_{p}^{ u_{p}}}{\prod\limits_{r} y_{r}^{ u_{r}}}\left(\frac{p}{p_{0}} ight)^{\sum\limits_{p} v_{p}-\sum\limits_{r} v_{r}} [/latex] (14.50)

[latex]\begin{aligned}&K=\frac{y_{ Cs ^{+} y_{ e }^{-}}}{y_{ Cs }}\left(\frac{p}{p_{0}} ight)^{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}[/latex]

TEST Analysis
The IGE model currently does not have [latex] Cs ^{+} [/latex] 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).

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

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