Behavior of Ideal Gases

The kinetic theory of gases postulates that the gas is composed of a very large number of particles called molecules. For an ideal gas, the volume of these molecules is insignificant compared with the total volume occupied by the gas. It is also assumed that these mole- cules have no attractive or repulsive forces between them, and it is assumed that all colli- sions of molecules are perfectly elastic.

Based on this kinetic theory of gases, a mathematical equation, called equation of state, can be derived to express the relationship existing between pressure, p, volume, V, and temperature, T,for a given quantity of moles of gas, n. This relationship for perfect gases, called the ideal gas law, is expressed mathematically by the following equation:

pV= nRT (3–1)

where

p= absolute pressure, psia V= volume, ft³

T = absolute temperature, °R n= number of moles of gas, lb-mole

R= the universal gas constant that, for these units, has the value 10.73 psia ft³/lb-mole °R

The number of pound-moles of gas, that is, n, is defined as the weight of the gas, m, divided by the molecular weight, M, or

(3–2) Combining equation (3–1) with (3–2) gives

(3–3) where

m= weight of gas, lb

M= molecular weight, lb/lb-mole

Since the density is defined as the mass per unit volume of the substance, equation (3–3) can be rearranged to estimate the gas density at any pressure and temperature:

(3–4) where ρ_g= density of the gas, lb/ft³.

It should be pointed out that lb refers to pounds massin all the subsequent discussions of density in this text.

ρ_g m V

pM

= = RT

pV m

M RT

= ⎛⎝⎜ ⎞

⎠⎟

n m

= M

EXAMPLE 3–1

Three pounds of n-butane are placed in a vessel at 120°F and 60 psia. Calculate the vol- ume of the gas assuming an ideal gas behavior.

SOLUTION

Step 1 Determine the molecular weight of n-butane from Table 1–1 to give M = 58.123

Step 2 Solve equation (3–3) for the volume of gas:

EXAMPLE 3–2

Using the data given in the preceding example, calculate the density of n-butane.

SOLUTION

Solve for the density by applying equation (3–4) :

Petroleum engineers usually are interested in the behavior of mixtures and rarely deal with pure component gases. Because natural gas is a mixture of hydrocarbon components, the overall physical and chemical properties can be determined from the physical proper- ties of the individual components in the mixture by using appropriate mixing rules.

The basic properties of gases commonly are expressed in terms of the apparent molec- ular weight, standard volume, density, specific volume, and specific gravity. These proper- ties are defined as in the following sections.

Apparent Molecular Weight, M_a

One of the main gas properties frequently of interest to engineers is the apparent molecu- lar weight. If y_irepresents the mole fraction of the ith component in a gas mixture, the apparent molecular weight is defined mathematically by the following equation.

(3–5) where

M_a= apparent molecular weight of a gas mixture

M_i= molecular weight of the ith component in the mixture y_i= mole fraction of component iin the mixture

Conventionally, natural gas compositions can be expressed in three different forms:

mole fraction, y_i, weight fraction, w_i, and volume fraction, v_i. M_a y M_i

i

= i

∑

= 1

ρg =( )( . )= ( . )(60 58 123) .

10 73 580 0 56lb/ft³ ρ_g m

V pM

= = RT V = ⎛⎝⎜ ⎞

⎠⎟

+ =

3 58 123

10 73 120 460

60 5 35 ³

.

( . )( )

. ft

V m

M RT

= ⎛⎝⎜ ⎞ p

⎠⎟

The mole fraction of a particular component, i, is defined as the number of moles of that component, n_i, divided by the total number of moles, n, of all the components in the mixture:

The weight fraction of a particular component, i, is defined as the weight of that com- ponent, m_i, divided by the total weight of m, the mixture:

Similarly, the volume fraction of a particular component, v_i, is defined as the volume of that component, V_i, divided by the total volume of V, the mixture:

It is convenient in many engineering calculations to convert from mole fraction to weight fraction and vice versa. The procedure is given in the following steps.

1. Since the composition is one of the intensive properties and independent of the quantity of the system, assume that the total number of gas is 1; that is, n= 1.

2. From the definitions of mole fraction and number of moles (see equation 3–2),

3. From the above two expressions, calculate the weight fraction to give

4. Similarly,

Standard Volume, V_sc

In many natural gas engineering calculations, it is convenient to measure the volume occu- pied by l lb-mole of gas at a reference pressure and temperature. These reference condi- tions are usually 14.7 psia and 60°F and are commonly referred to as standard conditions.

The standard volume then is defined as the volume of gas occupied by 1 lb-mole of gas at standard conditions. Applying these conditions to equation (3–1) and solving for the vol- ume, that is, the standard volume, gives

y w M

i w M

i i

i i

i

=

∑

^//

w m

m m

m

y M y M

y M

i M

i i

i i

i i

i i

i

i i

a

= = = =

∑ ∑

m_i =n M_{i i} = y M_{i i}

y n

n

n n

i

i i

= = = i

1

v V

V V

i V

i i

i i

= =

∑

w m

m m

i m

i i

i i

= =

∑

y n

n n

i n

i i

i i

= =

∑

V_sc= 379.4 scf/lb-mole (3–6) where

V_sc= standard volume, scf/lb-mole scf = standard cubic feet

T_sc= standard temperature, °R p_sc= standard pressure, psia Gas Density, ^ρρ_g

The density of an ideal gas mixture is calculated by simply replacing the molecular weight, M, of the pure component in equation (3–4) with the apparent molecular weight, M_a, of the gas mixture to give

(3–7) where ρ_g= density of the gas mixture, lb/ft³, and M_a= apparent molecular weight.

Specific Volume, v

The specific volume is defined as the volume occupied by a unit mass of the gas. For an ideal gas, this property can be calculated by applying equation (3–3):

(3–8) where v= specific volume, ft³/lb, and ρ_g= gas density, lb/ft³.

Specific Gravity,

γγ

_g

The specific gravity is defined as the ratio of the gas density to that of the air. Both densities are measured or expressed at the same pressure and temperature. Commonly, the standard pressure, p_sc, and standard temperature, T_sc, are used in defining the gas specific gravity.

(3–9) Assuming that the behavior of both the gas mixture and the air is described by the ideal gas equation, the specific gravity can be then expressed as

or

(3–10)

γ_g M^{a a}

M

= = M

air 28 96. γ_g

p Ma

RT p M

RT

=

sc sc sc air

sc

γ_g =gas density @ 14.7 and 60°

air density @ 114.7 and 60° = ρ_air ρ

g

v V m

RT pM_{a g}

= = = 1

ρ ρ_g pM^a

= RT

sc sc

sc

(1) (1)(10.73)(520) V = RT 14.7

p =

where

γ_g= gas specific gravity, 60°/60°

ρ_air= density of the air

M_air= apparent molecular weight of the air = 28.96 M_a= apparent molecular weight of the gas

p_sc= standard pressure, psia T_sc= standard temperature, °R

EXAMPLE 3–3

A gas well is producing gas with a specific gravity of 0.65 at a rate of 1.1 MMscf/day. The average reservoir pressure and temperature are 1500 psi and 150°F. Calculate

1. The apparent molecular weight of the gas.

2. The gas density at reservoir conditions.

3. The flow rate in lb/day.

SOLUTION

From equation (3–10), solve for the apparent molecular weight:

M_a= 28.96 γ_g

M_a= (28.96)(0.65) = 18.82

Apply equation (3–7) to determine gas density:

The following steps are used to calculate the flow rate.

Step 1 Because 1 lb-mole of any gas occupies 379.4 scf at standard conditions, then the daily number of moles, n, that the gas well is producing is

Step 2 Determine the daily mass, m, of the gas produced from equation (2–2):

m = nM_a

m = 2899(18.82) = 54,559 lb/day

EXAMPLE 3–4

A gas well is producing a natural gas with the following composition:

COMPONENT y_i

CO₂ 0.05

C₁ 0.90

C₂ 0.03

C₃ 0.02

Assuming an ideal gas behavior, calculate the apparent molecular weight, specific gravity, gas density at 2000 psia and 150°F, and specific volume at 2000 psia and 150°F.

n=( . )( ) = . 1 1 10

379 4 2899

6

lb-moles ρ_g=( )( . )=

( . )(1500 18 82) . 10 73 610 4 31lb/ft³

SOLUTION

The table below describes the gas composition.

Component y_i M_i y_iM_i

CO₂ 0.05 44.01 2.200

C₁ 0.90 16.04 14.436

C₂ 0.03 30.07 0.902

C₃ 0.02 44.11 0.882

M_a= 18.42

Apply equation (3–5) to calculate the apparent molecular weight:

M_a= 18.42

Calculate the specific gravity by using equation (3–10):

γ_g= M_a/28.96= 18.42/28.96 = 0.636

Solve for the density by applying equation (3–7):

Determine the specific volume from equation (3–8):