A gas is found to have the following composition. Component Composition, mole fr

Question

A gas is found to have the following composition.

Component Composition, mole fraction

Methane 0.6904

Ethane 0.0864

Propane 0.0534

i-Butane 0.0115

n-Butane 0.0233

i-Pentane 0.0093

n-Pentane 0.0085

Hexanes 0.0173

Heptanes plus 0.0999

Properties of heptanes plus

Specific gravity 0.827

Molecular weight 158 lb/lb mole

1) Calculate the specific gas gravity

2) Calculate the molar volume if the reservoir temperature is 90°F and reservoir pressure is

455 psia. Then compare your result with the molar volume of an ideal gas at the same

conditions.

3) Calculate the mass of gas in the reservoir at initial conditions. Give your answer in lb

moles if this gas is a retrograde gas. The discovery pressure in reservoir, 7000 psig, is

higher than dew point pressure, 6010 psig. Reservoir temperature is 256°F. The reservoir

lies under 7040 acres, has an average thickness of 13 ft, has a porosity of 11 %, and has a

water saturation of 40%.

Explanation / Answer

1. To find the specific gravity first of all we need to find the average molar mass of the gas mixture.form the following formulae

M(gas mixture) = Summation of [YiMi],

where Yi = mole fraction of a gaseous component

Mi = Molar mass of the gaseous component.

For methane, Yi = 0.6904, Mi = 16.0 g/mol

=> YiMi = 0.6904 x 16.04 g/mol = 11.074 g/mol

For ethane, Yi = 0.0864, Mi = 30.07 g/mol

=> YiMi = 0.0864 x 30.07 g/mol = 2.598 g/mol

For propane, Yi = 0.0534, Mi = 44.1 g/mol

=> YiMi = 0.0534 x 44.1 g/mol  = 2.355 g/mol

For i-butane, Yi = 0.0115, Mi = 58.12 g/mol

=> YiMi = 0.0115x58.12 g/mol = 0.668 g/mol

For n-butane, Yi = 0.0233, Mi = 58.12 g/mol

=> YiMi = 0.0233x58.12  g/mol = 1.354 g/mol

For i-pentane, Yi = 0.0093, Mi = 72.15 g/mol

=> YiMi = 0.0093x72.15 g/mol = 0.671 g/mol

For n-pentane, Yi = 0.0085, Mi = 72.15 g/mol

=> YiMi = 0.0085x72.15 g/mol = 0.613 g/mol

For n-hexane, Yi = 0.0173, Mi = 86.18 g/mol

=> YiMi = 0.0173x86.18 g/mol = 1.491 g/mol

For heptane plus, Yi = 0.0173, Mi = 158 lb/lb mol = 158 g/mol

=> YiMi = 0.0999x158 g/mol = 15.784 g/mol

M(gas mixture) = Summation of [YiMi]

= 11.074 g/mol + 2.598 g/mol + ,..........+ 15.784 g/mol = 36.6 g/mol

Molecular mass of air, Ma = 28.976 g/mol

Hence specific gravity = Mg / Ma = 36.6 g/mol / 28.967 g/mol = 1.263 (answer)

2. n = 1 mol, T = 305.4 K

P = 455 psia = 30.96 atm

Molar volume = nRT / P = 1 x 0.0821L.atm.mol-1K-1x 305.4 K / 30.96 atm = 0.810 L (answer)

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