# 5.2: Ideal Gas Behavior

For an $$N$$-component ideal gas mixture we have the following relations

$p_{A} V = n_{A} RT , \quad A = 1, 2, ...N \label{2}$

Here $$p_{A}$$ is the partial pressure of species $$A$$ and $$R$$ is the gas constant. Values of the gas constant in different units are given in Table $$\PageIndex{1}$$. If we sum Equation \ref{2} over all $$N$$ species we obtain

$pV = nRT \label{3}$

where

$p = \sum_{A = 1}^{A = N}p_{A} \label{4}$

$n = \sum_{A = 1}^{A = N}n_{A} \label{5}$

Equations \ref{2} through \ref{5} are sometimes referred to as Dalton’s Laws.

Table $$\PageIndex{1}$$. Numerical values of the gas constant, $$R$$
Numerical value Units
8.314 m$$^{3}$$ Pa/ mol K
8.314 J/mol K
0.08314 liter bar/mol K
82.06 atm-cm$$^{3}$$ /mol K
1.986 cal/mol K

In Figure $$\PageIndex{1}$$ we have illustrated a constant pressure, isothermal mixing process. In the compartment containing species $$A$$ illustrated in Figure $$\PageIndex{1a}$$, we can use Equation \ref{2} to obtain

$p V_{A} = n_{A} RT \label{6}$

while in the compartment containing species $$B$$ we have

$p V_{B} = n_{B} RT \label{7}$

Upon removal of the partition and mixing, we have the situation illustrated in Figure $$\PageIndex{1b}$$. For that condition, we can use Equation \ref{3} to obtain

$pV = (n_{A} +n_{B} )RT \label{8}$

where the volume is given by

$V = V_{A} + V_{B} + \Delta V_{ mix} \label{9}$

By definition, an ideal gas mixture obeys what is known as Amagat’s Law, i.e.

$\Delta V_{ mix} = 0 , \quad \text{ Amagat's Law} \label{10}$

Amagat’s law can also be expressed as

$\sum_{A = 1}^{A = N}V_{A} = V , \quad \text{ Amagat's Law} \label{11}$

In the gas phase the mole fraction is generally denoted by $$y_{A}$$ while $$x_{A}$$ is reserved for mole fractions in the liquid phase. For ideal gas mixtures it is easy to show, using Eqs. $$(5.1.6)$$, \ref{2} and \ref{3}, that the mole fraction is given by

$y_{A} = p_{A} / p \label{12}$

This is an extremely convenient representation of the mole fraction in terms of the partial pressure; however, one must always remember that it is strictly valid only for an ideal gas.

Example $$\PageIndex{1}$$: Flow of an ideal gas in a pipeline

A large pipeline is used to transport natural gas from Oklahoma to Nebraska as illustrated in Figure $$\PageIndex{2}$$.

Natural gas, consisting of methane with small amounts of ethane, propane, and other low molecular mass hydrocarbons, can be assumed to behave as an ideal gas at ambient temperature. At the pumping station in Glenpool, OK, the pressure in the 20-inch pipeline is 2900 psia and the temperature is $$T_o = 90$$ F. At the receiving point in Lincoln, NE, the pressure is 2100 psia and the temperature is $$T_1 = 45$$ F. The mass average velocity of the gas at Glenpool is 50 ft/s. Assuming ideal gas behavior and treating the gas as 100% methane, compute the following:

1. Mass averaged velocity at the end of the pipeline
2. Mass and molar flow rates at both ends of the pipeline.

This problem has been presented in terms of a variety of units, and it is often convenient to express all variables in terms of SI units as follows:

$(2900 \text{ psia}) \times \frac{6895 \ Pa}{\text{psia}} \times \frac{MPa}{10^{6} \ Pa}=20.00 \ MPa \nonumber$

$(2100 \text{ psia}) \times \frac{6895 \ Pa}{\text{psia}} \times \frac{MPa}{10^{6} \ Pa}=14.48 \ MPa \nonumber$

$(90 \ F-32 \ F) \times \frac{5 \ C}{9 \ F} \times \frac{K}{C}+273.15 \ {K}=305.4 \ {K}\nonumber$

$(45 \ F-32 \ F) \times \frac{5 \ C}{9 \ F} \times \frac{K}{C}+273.15 \ {K}=280.4 \ {K} \nonumber$

$(50 \ {ft}/{s}) \times \frac{0.3048 \ m}{ft}=15.24 \ {m}/{s} \nonumber$

$D_{o}=D_{1}=(20 \ {in}) \times \frac{0.0254 \ m}{in}=0.508 \ {m} \nonumber$

$A_o = A_1 = \frac{\pi D^{2} }{4} = \frac{\pi \left({ 0.508} \ { m}\right)^{2} }{4} = 0.2027 \ { m}^{2}\nonumber$

Since the natural gas is assumed to be pure methane, we can obtain the molecular mass from Table A2 in Appendix A where we find $$MW_{\ce{CH4}} = 16.043$$ g/mol. The volume per mole of methane can be determined using the ideal gas law given by Equation \ref{2} along with the value of the gas constant found in Table $$\PageIndex{1}$$. The volume per mole at the entrance is determined as

${V_o / n} = \frac{R T}{p_o } = \frac{8.314 \frac{ m^{ 3} \ Pa}{ mol \ K} \left(305.4 \ K \right)}{19.98\times 10^{6} \ { Pa}} = 1.2\times 10^{-4} \ { m}^{ 3} /{ mol} \tag{1} \label{a}$

while the volume per mole at the exit is given by

${V_1 / n} = \frac{ R T}{p_{1} } = 1.53\times 10^{-4} \ { m}^{ 3} /{ mol} \label{b}\tag{2}$

The gas densities at the entrance and exit of the pipeline are given by

$(\rho_{\ce{CH4}} )_o = \frac{MW_{\ce{CH4}} }{V_{o} / n } = \frac{16.043 \ g/mol}{1.2\times 10^{-4} \ m^{ 3}/mol} = 133,700 \ g/m^{ 3} = 133.7 \ kg/m^{ 3} \label{c}\tag{3}$

$(\rho_{\ce{CH4}} )_1 = \frac{MW_{\ce{CH4}} }{V_{1} / n } = 104.8 \ kg/m^{ 3}\label{d}\tag{4}$

To perform a mass balance for the pipeline, we begin with the species mass balance given by

$\frac{d}{dt} \int_{\mathscr{V}}\rho_{A} dV + \int_{\mathscr{A}}(\rho_{A} \mathbf{v}_A ) \cdot \mathbf{n} dA = \int_{\mathscr{V}}r_{A} dV \label{e}\tag{5}$

Since there are no chemical reactions and there is no accumulation, the first and last terms in this result are zero and Equation \ref{e} simplifies to

$\int_{\mathscr{A}}(\rho_{A} \mathbf{v}_A ) \cdot \mathbf{n} dA = 0 \label{f}\tag{6}$

For a single component system, the species velocity is equal to the mass average velocity (see Chap. 4) and Equation \ref{f} takes the form

$\int_{\mathscr{A}}\rho_{A} \mathbf{v}\cdot \mathbf{n} dA = 0 \label{g}\tag{7}$

The control volume is constructed in the obvious manner, thus there is an entrance in Glenpool, OK and an exit in Lincoln, NE. Since the mass average velocity and the diameter of the pipeline are given, we express the mass balance as

$(\rho_{\ce{CH4}} )_o { v}_o A_o = (\rho_{\ce{CH4}} )_1 { v}_1 A_1 \label{h} \tag{8}$

The only unknown in this result is the velocity of the gas at the exit of the pipeline. Solving for $${ v}_1$$ we obtain

${ v}_1 = \left[ \frac{A_o }{A_1 } \frac{(\rho_{\ce{CH4}} )_o }{(\rho_{\ce{CH4}} )_1 } \right] { v}_o = \frac{133.7 \ kg/m^{ 3} }{104.8 \ kg/m^{ 3} } \left(15.24 \ {m/s} \right) = 19.44 \ {m/s} \label{i}\tag{9}$

The mass flow rate is a constant given by

$\dot{m}_o = \dot{m}_1 = (\rho_{\ce{CH4}} )_o { v}_o A_{ o} = \left(133.7 \ kg/m^{ 3} \right) \left(15.24 \ {m/s}\right) \left(0.2027 \ { m}^{ 2} \right) = 413 \ kg/s \label{j}\tag{10}$

from which we determine the constant molar flow rate to be

$\dot{M}_o = \dot{M}_1 = \frac{(\dot{m}_{\ce{CH4}} )_o }{MW_{\ce{CH4}} } = \frac{413 \ kg/s}{16.043 \ g/mol} = 25.74\times 10^{3} \ mol/s \label{k}\tag{11}$

In order to use Equation \ref{3} to estimate the density of a pure gas, we multiply by the molecular mass and arrange the result in the form

$\rho = \frac{n MW}{V} = \frac{p MW}{R T} \label{13}$

For an ideal gas mixture, one uses the definitions of the total mass density of a mixture (Equation $$(5.1.2)$$) and total pressure (Equation \ref{4}) along with Equation \ref{2} to obtain

$\rho = \sum_{A = 1}^{A = N}\rho_{A} = \sum_{A = 1}^{A = N}\frac{n_{A} MW_{A} }{V} = \sum_{A = 1}^{A = N}\frac{p_{A} MW_{A} }{R T} = \frac{p \overline{MW}}{R T} \label{14}$

Here we have used $$\overline{MW}$$ to represent the molar average molecular mass defined by

$\overline{MW} = \frac{1}{p} \sum_{A = 1}^{A = N}p_{A} MW_{A} = \sum_{A = 1}^{A = N}y_{A} MW_{A} \label{15}$

These results are applicable when molecule-molecule interaction is negligible and this occurs for many gases under ambient conditions. At low temperatures and high pressures, gases depart from ideal gas behavior, and under those conditions one should use more accurate equations of state, such as those presented in a standard course on thermodynamics1.

Example $$\PageIndex{2}$$: Molecular mass of air

The air we breathe has a composition that depends on position. Air pollution sources abound and these sources add minute amounts of chemicals to the atmosphere. Combustion of fuels in cars and power plants are a source for sulfur dioxide, oxides of nitrogen, and carbon monoxide. Chemical industries add pollutants such as ammonia, chlorine, and even hydrogen cyanide to the atmosphere. In some locations, there are minute amounts of other gases such as argon, helium, and radon. Standard dry air, for the purpose of combustion computations, is assumed to be a mixture of 79% by volume of nitrogen and 21% by volume of oxygen. In this example, we want to determine the molar average molecular mass of standard dry air.

For an ideal gas, the volume percentage is also the molar percentage. Thus, the volume percentages of nitrogen and oxygen can be simply translated to mole fractions according to

79% by volume of nitrogen $$\rightarrow y_{\ce{N_2}} = 0.79$$

21% by volume of oxygen $$\rightarrow y_{\ce{O_2}} = 0.21$$

We can use Equation \ref{15} to compute the molar average molecular mass of standard dry air according to

$\overline{MW} = MW_{\text{dry air}} = y_{\ce{N2}} MW_{\ce{N2}} + y_{\ce{O2}} MW_{\ce{O2}} = 28.85 \ g/mol \nonumber$