/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 76 Determining the value of newly l... [FREE SOLUTION] | 91Ó°ÊÓ

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Determining the value of newly located natural gas sites involves estimating the gas composition. quantity, and ease of access. For example, one report described a find of 2 trillion cubic feet of natural gas that is significantly offshore, in 20 feet of water, and at a drilled depth of 25,000 ft. (In North America and the OPEC countries, reported volumes are determined at 14.73 psia and \(60^{\circ} \mathrm{F}\).) The pressure in this find is estimated to be 750 atm, and the gas is 94 mole \(\%\) methane, \(3.5 \%\) ethane, and the balance \(\mathrm{CO}_{2}\) (a) Estimate the total Ib-moles of gas in the find. (b) Use the compressibility-factor equation of state to estimate the specific volume (ft \(^{3} /\) /b-mole) in the well. The temperature of such wells can vary depending upon a number of factors; for the purposes of this problem, assume that it is \(200^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The estimated total lb-moles of gas in the find and specific volume in the well can only be obtained after carrying out the above calculations. Particular values depend upon the provided compressibility factors for each gas at given conditions.

Step by step solution

01

Conversion to lb-moles

First, it is important to convert the volume of gas to lbs-moles. To accomplish this, one can use the ideal gas law \(PV = nRT\), where \(P = 14.73\) psia (pressure), \(V = 2\) trillion cubic feet (volume), \(R = 10.73\) ft³.psi/(R.lb-mol) (universal gas constant) and \(T = 520\)R (temperature in Rankine, instrumental in getting the gas volumes to base conditions at \(60^{\circ} \mathrm{F}\)). Solving for \(n\) (the amount in mole) will give the total lb-moles of gas in the find.
02

Compressibility Calculation

The second step is to calculate the specific volume in the well using the compressibility-factor equation of state. Assuming the system behaves ideally, the compressibility factor \(Z\), which represents how closely real gases obey the ideal gas law, should be one. However, due to deviations at high pressure and temperature, the overall \(Z\) has to be calculated based on mole fractions. The fractions are 94 mole% methane, 3.5% ethane, and the balance CO2. Thus, the overall compressibility at 200°C (~673°R) and 750 atm can be estimated by summing up the product of each component's mole fraction and its respective compressibility factor under these conditions.
03

Specific Volume Calculation

Having calculated the overall compressibility factor, the gas specific volume at 750 atm (around 11265 psia) and 200°C can be calculated from the equation \(v = ZRT/P\), where \(R = 10.73\) ft³.psi/(R.lb-mol), \(T = 200°C = 673\)R, and \(P = 750 atm = 11265 psia\). This yields the specific volume in ft³/lb-mole.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Ideal Gas Law and its Application
The ideal gas law is a fundamental equation in the study of thermodynamics, particularly in understanding the behavior of gases under various conditions. The equation is expressed as \(PV = nRT\), which relates the pressure (P), volume (V), and temperature (T) of a gas to the amount of substance (n) in moles, and includes the ideal gas constant (R). In practical terms, this law helps us to calculate the quantity of a gas, like in our exercise where we have to find the total lb-moles of gas in a newly discovered natural gas find.

When applying the ideal gas law, it is critical to ensure that the units used are consistent. For instance, the pressure must be in absolute terms, such as psi (pounds per square inch absolute), and temperature should be in a universal scale like Kelvin (K) or Rankine (R). Any volume must be in cubic units consistent with the gas constant used. In our problem, we work with psia for pressure, cubic feet for volume, and Rankine for temperature, while the gas constant is given in ft³.psi/(R.lb-mol).

It's important to note that the ideal gas law assumes gases are 'ideal', meaning they have no intermolecular forces and occupy no space; however, real gases only approximate this behavior under certain conditions, usually at low pressures and high temperatures. When dealing with high-pressure conditions like those in gas wells, we must account for the real nature of gas, which leads us to the compressibility-factor equation of state.
Real Gas Behavior and the Compressibility-Factor Equation of State
As gases deviate from ideal behavior, adjustments have to be made to account for the interactions and volume occupied by gas molecules. This is where the compressibility-factor (Z) comes into play. The compressibility-factor equation of state expresses how much the real gas deviates from the predicted behavior of an ideal gas and is crucial in high-pressure and high-temperature environments, like those found in deep gas wells.

Compressibility Factor (Z)

The compressibility factor is a dimensionless value that is multiplied by the ideal gas law to correct for non-ideal behavior. It is defined by the relationship \(Z = \frac{PV}{nRT}\), where a Z-value of 1 means the gas behaves ideally. For real gases, Z can vary from 1 depending on factors like pressure, temperature, and gas composition.

In the exercise, we use this concept to estimate the specific volume (ft³/lb-mole) in the well. Given that the well has a very high pressure and a high temperature, we expect significant deviations from ideality. To estimate the compressibility factor of the gas mixture in the well, we combine the compressibility factors of each component, weighted by their mole fractions, to calculate an overall Z for the mixture. Here, we consider the mole percentage of methane, ethane, and carbon dioxide, which affects their individual and collective behavior under the described conditions.
Mole Fraction Calculation and its Importance in Gas Mixtures
In chemistry and engineering, mole fraction is a way of expressing the concentration of a component in a mixture. The mole fraction (chi, represented by the Greek letter \(\chi\)) of any component is defined as the number of moles of that component divided by the total number of moles of all components in the mixture.

Calculating Mole Fractions

To calculate the mole fraction, one would use the formula: \(\chi_i = \frac{n_i}{n_{total}}\), where \(\chi_i\) is the mole fraction of component i, \(n_i\) is the number of moles of component i, and \(n_{total}\) is the total number of moles in the mixture.

Mole fractions are essential when working with gas mixtures as they help us determine the behavior of each gas component within the mixture. In our exercise involving a natural gas find, understanding the mole fractions of methane, ethane, and carbon dioxide is critical for calculating the overall compressibility factor of the gas mixture. These fractions are factored into the equation when assessing how close the mixture is to ideal behavior, thereby enabling us to accurately adjust the gas laws to fit the reality of the mixture in the high-pressure and high-temperature conditions of the gas well.

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Most popular questions from this chapter

During your summer vacation, you plan an epic adventure trip to scale Mt. Kilimanjaro in Tanzania. Dehydration is a great danger on such a climb, and it is essential to drink enough water to make up for the amount you lose by breathing. (a) During your pre-trip physical, your physician measured the average flow rate and composition of the gas you exhaled (expired air) while performing light activity. The results were \(11.36 \mathrm{L} / \mathrm{min}\) at body temperature (37^) C) and 1 atm, 17.08 mole\% oxygen, 3.25\% carbon dioxide, 6.12 mole\% \(\mathrm{H}_{2} \mathrm{O},\) and the balance nitrogen. The ambient (inspired) air contained 1.67 mole\% water and a negligible amount of carbon dioxide. Calculate the rate of mass lost through the breathing process (kg/day) and the volume of water in liters you would have to drink per day just to replace the water lost in respiration. Consider your lungs to be a continuous steady-state system, with input streams being inspired air and water and \(\mathrm{CO}_{2}\) transferred from the blood and output streams being expired air and \(\mathrm{O}_{2}\) transferred to the blood. Assume no nitrogen is transferred to or from the blood. (b) You made the trip to Tanzania and completed the climb to Uhuru Peak, the summit of Kilimanjaro, at an altitude of 5895 meters above sea level. The ambient temperature and pressure there averaged \(-9.4^{\circ} \mathrm{C}\) and \(360 \mathrm{mm} \mathrm{Hg},\) and the air contained \(0.46 \mathrm{mole} \%\) water. The molar flow rate of your expired air was roughly the same as it had been at sea level, and the expired air contained \(14.86 \% \mathrm{O}_{2}\) \(3.80 \% \mathrm{CO}_{2},\) and \(13.20 \% \mathrm{H}_{2} \mathrm{O} .\) Calculate the rate of mass lost (g/day) through breathing and water you would have to drink (L/day) just to replace the water lost in respiration. (c) The equality of the molar flow rates of expired air at sea level and at Uhuru Peak is due to a cancellation of effects, one of which would tend to increase the rate at higher altitudes and the other to decrease it. What are those effects? (Hint: Use the ideal-gas equation of state in your solution, and think about how the oxygen concentration at a high altitude would likely affect your breathing rate.)

A gas consists of 20.0 mole \(\% \mathrm{CH}_{4}, 30.0 \% \mathrm{C}_{2} \mathrm{H}_{6},\) and \(50.0 \% \mathrm{C}_{2} \mathrm{H}_{4} .\) Ten kilograms of this gas is to be compressed to a pressure of 200 bar at \(90^{\circ} \mathrm{C}\). Using Kay's rule, estimate the final volume of the gas.

Most of the concrete used in the construction of buildings, roads, dams, and bridges is made from Portland cement, a substance obtained by pulverizing the hard, granular residue (clinker) from the roasting of a mixture of clay and limestone and adding other materials to modify the setting properties of the cement and the mechanical properties of the concrete. The charge to a Portland cement rotary kiln contains \(17 \%\) of a dried building clay \(\left(72 \mathrm{wt} \% \mathrm{SiO}_{2}\right.\) \(\left.16 \% \mathrm{Al}_{2} \mathrm{O}_{3}, 7 \% \mathrm{Fe}_{2} \mathrm{O}_{3}, 1.7 \% \mathrm{K}_{2} \mathrm{O}, 3.3 \% \mathrm{Na}_{2} \mathrm{O}\right)\) and \(83 \%\) limestone \(\left(95 \mathrm{wt} \% \mathrm{CaCO}_{3}, 5 \% \text { impuritics }\right)\) When the solid temperature reaches about \(900^{\circ} \mathrm{C},\) calcination of the limestone to lime (CaO) and carbon dioxide occurs. As the temperature continues to rise to about \(1450^{\circ} \mathrm{C},\) the lime reacts with the minerals in the clay to form such compounds as \(3 \mathrm{CaO} \cdot \mathrm{SiO}_{2}, 3 \mathrm{CaO} \cdot \mathrm{Al}_{2} \mathrm{O}_{3},\) and \(4 \mathrm{CaO} \cdot \mathrm{Al}_{2} \mathrm{O}_{3} \cdot \mathrm{Fe}_{2} \mathrm{O}_{3} .\) The flow rate of \(\mathrm{CO}_{2}\) from the kiln is \(1350 \mathrm{m}^{3} / \mathrm{h}\) at \(1000^{\circ} \mathrm{C}\) and 1 atm. Calculate the feed rates of clay and limestone ( \(\mathrm{kg} / \mathrm{h}\) ) and the weight percent of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the final cement.

Phosgene (CCl, O) is a colorless gas that was used as an agent of chemical warfare in World War I. It has the odor of new-mown hay (which is a good warning if you know the smell of new-mown hay). Pete Brouillette, an innovative chemical engincering student, came up with what he believed was an effective new process that utilized phosgene as a starting material. He immediately set up a reactor and a system for analyzing the reaction mixture with a gas chromatograph. To calibrate the chromatograph (i.e., to determine its response to a known quantity of phosgene), he evacuated a 15.0 cm length of tubing with an outside diameter of \(0.635 \mathrm{cm}\) and a wall thickness of \(0.559 \mathrm{mm}\), and then connected the tube to the outlet valve of a cylinder containing pure phosgene. The idea was to crack the valve, fill the tube with phosgene, close the valve, feed the tube contents into the chromatograph, and observe the instrument response. What Pete hadn't thought about (among other things) was that the phosgene was stored in the cylinder at a pressure high enough for it to be a liquid. When he opened the cylinder valve, the liquid rapidly flowed into the tube and filled it. Now he was stuck with a tube full of liquid phosgene at a pressure the tube was not designed to support. Within a minute he was reminded of a tractor ride his father had once given him through a hayfield, and he knew that the phosgene was leaking. He quickly ran out of the lab, called campus security, and told them that a toxic leak had occurred, that the building had to be evacuated, and the tube removed and disposed of properly. Personnel in air masks shortly appeared, took care of the problem, and then began an investigation that is still continuing. (a) Show why one of the reasons phosgene was an effective weapon is that it would collect in low spots soldiers often mistakenly entered for protection. (b) Pete's intention was to let the tube equilibrate at room temperature ( \(23^{\circ} \mathrm{C}\) ) and atmospheric pressure. How many gram-moles of phosgene would have been contained in the sample fed to the chromatograph if his plan had worked? (c) The laboratory in which Pete was working had a volume of \(2200 \mathrm{ft}^{3}\), the specific gravity of liquid phosgene is \(1.37,\) and Pete had read somewhere that the maximum "safe" concentration of phosgene in air is \(0.1 \mathrm{ppm}\) \(\left(0.1 \times 10^{-6} \mathrm{mol} \mathrm{CCl}_{2} \mathrm{O} / \mathrm{mol}\) air) \right. Would the "safe" concentration have been exceeded if all the liquid phosgene in the tube had evaporated into the room? Even if the limit would not have been exceeded, give several reasons why the lab would still have been unsafe. (d) List several things Pete did (or failed to do) that made his experiment unnecessarily hazardous.

Hydrogen sulfide has the distinctive unpleasant odor associated with rotten eggs, and it is poisonous. It often must be removed from crude natural gas and is therefore a product of refining natural gas. In such instances, the Claus process provides a means of converting \(\mathrm{H}_{2} \mathrm{S}\) to elemental sulfur. Consider a feed stream to a Claus process that consists of 10.0 mole \(\% \mathrm{H}_{2} \mathrm{S}\) and \(90.0 \% \mathrm{CO}_{2}\). Onethird of the stream is sent to a furnace where the \(\mathrm{H}_{2} \mathrm{S}\) is burned completely with a stoichiometric amount of air fed at 1 atm and \(25^{\circ} \mathrm{C}\). The combustion reaction is $$\mathrm{H}_{2} \mathrm{S}+\frac{3}{2} \mathrm{O}_{2} \rightarrow \mathrm{SO}_{2}+\mathrm{H}_{2} \mathrm{O}$$ The product gases from this reaction are then mixed with the remaining two- thirds of the feed stream and sent to a reactor in which the following reaction goes to completion: $$2 \mathrm{H}_{2} \mathrm{S}+\mathrm{SO}_{2} \rightarrow 3 \mathrm{S}+2 \mathrm{H}_{2} \mathrm{O}$$ The gases leave the reactor at \(10.0 \mathrm{m}^{3} / \mathrm{min}, 320^{\circ} \mathrm{C},\) and \(205 \mathrm{kPa}\) absolute. Assuming ideal-gas behavior, determine the feed rate of air in kmol/min. Provide a single balanced chemical equation reflecting the overall process stoichiometry. How much sulfur is produced in \(\mathrm{kg} / \mathrm{min} ?\)

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