Equilibrium of an Ideal Gas Mixture A natural gas reservoir contains a mixture of methane and nitrogen at 298 K. At the top of the well, the mixture pressure is 100 kPa and the molar composition is found to be 50% methane and 50% nitrogen. Determine (a) the pressure and (b) the molar composition

Question

Equilibrium of an Ideal Gas Mixture

A natural gas reservoir contains a mixture of methane and nitrogen at 298 K. At the top of the well, the mixture pressure is 100 kPa and the molar composition is found to be 50% methane and 50% nitrogen. Determine (a) the pressure and (b) the molar composition at the bottom, 3 km below the surface. Assume uniform temperature and ideal gas behavior.

Accepted Answer

Determine the partial pressure of each component at the bottom of the well, using Eq. (14.16), from which the molar composition and total pressure can be deduced.

[latex] \frac{p_{1}}{p_{2}}=\exp \left[\frac{g\left(z_{2}-z_{1} ight)}{(1000 J / kJ ) R T} ight] [/latex] (14.16)

Assumptions
Uniform temperature throughout the well. The IG mixture model is applicable for the gas mixture.

Analysis
From Table C-1, or any IG TESTcalc, we obtain [latex] R_{ CH _{4}} ext { as } 0.512 kJ / kg \cdot K ext { and } R_{ N _{2}} ext { as } 0.297 [/latex] kJ/kg ⋅ K. Representing the state of the gas at the bottom and top by State-1 and State-2, respectively, the component partial pressures can be evaluated at the two states as:

[latex]\begin{aligned}p_{ CH _{4}, 2}=y_{ CH _{4}, 2} p_{2} &=(0.5)(100)=50 kPa \p_{ N _{2}, 2} &=p_{2}-p_{ CH _{4}, 2}=50 kPa\end{aligned}[/latex]

Using Eq. (14.16),

[latex]\begin{aligned}p_{ CH _{4}, 1} &=p_{ CH _{4}, 2} \exp \left[\frac{g\left(z_{2}-z_{1} ight)}{(1000 J / kJ ) R_{ CH _{4}} T} ight]=(50) \exp \left[\frac{(9.81)(3000)}{(1000 J / kJ )(0.512)(298)} ight] \&=60.6 kPa\end{aligned}[/latex]

Similarly,

[latex]\begin{aligned}p_{ N _{4}, 1} &=(50) \exp \left[\frac{(9.81)(3000)}{(1000 J / kJ )(0.297)(298)} ight] \&=69.7 kPa\end{aligned}[/latex]

Therefore, the total pressure at the bottom, State-1, is:

[latex] p_{1}=p_{ CH _{4,1}}+p_{ N _{2,1}}=60.6+69.7=130.3 kPa [/latex]

And the molar composition is: [latex] y_{ CH _{4}, 1}=p_{ CH _{4}, 1} / p_{1}=60.6 / 130.3=0.465 [/latex]

[latex] y_{ N _{4}, 1}=1-p_{ CH _{4}, 1}=1-0.465=0.535 [/latex]

TEST Analysis
Launch the n-IG system-state TESTcalc. Add 1 kmol of methane as the only component. Set up State-2 with p2=50 kPa, z2=3000 m, and T2=298 K. Evaluate State-1, with T1=T2, z1=0, and g1=g2+9.81*z2/1000, yielding p1 as 60.5 kPa. Similarly, find the pressure of nitrogen by repeating the solution for pure nitrogen. Given that the two gases exist independently, the evaluated pressures are partial pressures from which the mole fractions can be evaluated.

Discussion
Methane and nitrogen coexist in Dalton’s model without any special intermolecular attraction or repulsion. An ideal solution can be similarly treated, with each component being in equilibrium independently. For example, under equilibrium condition, the concentration of salt can be expected to vary with depth in the oceans. However, the time required to reach equilibrium through diffusion is much longer than the time of mixing due to ocean currents; therefore, oceans have a remarkably constant concentration of salt at all depths.

Question 14.2: Equilibrium of an Ideal Gas Mixture A natural gas reservoir ...

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