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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.

Short Answer

Expert verified
The final volume is the value obtained from the calculation in Step 5. The procedure involves using Kay's rule to determine the pseudo-critical properties of the gas mixture, and then using these values in the Ideal Gas Law to determine the final volume.

Step by step solution

01

Finding Molar Fractions

First, convert the percentages into mole fractions. Mole fractions are equal to the given percentages divided by 100. Therefore for methane \(CH_4\), we get (20/100)= 0.2, for ethane \(C_2H_6\), (30/100) = 0.3 and for ethylene \(C_2H_4\), (50/100) = 0.5.
02

Calculation of Pseudo Critical Pressure

Calculation of the Pseudo critical pressure (Pc) and Pseudo critical temperature (Tc) by using Kay’s rule. Pseudo critical constants for methane, ethane and ethylene are: \(Pc(CH_4) = 46 bars, Pc(C_2H_6) = 48 bars, Pc(C_2H_4) = 50 bars, Tc(CH_4) = -82.6^{\circ}C, Tc(C_2H_6) =32.2^{\circ}C , Tc(C_2H_4) = 9.9^{\circ}C\). Multiply the mole fraction of each gas by its critical pressure to find the partial pressures. Then add all the partial pressures to obtain the pseudo critical pressure. \(Pc_{mix} = (Peercentage(CH_4) * Pc(CH_4)) + (Peercentage(C_2H_6) * Pc(C_2H_6)) + (Peercentage(C_2H_4) * Pc(C_2H_4)) \)
03

Calculation of Pseudo Critical Temperature

Multiply the mole fraction of each gas by its critical temperature to find the partial temperatures. Then add all the partial temperatures to obtain the pseudo critical temperature. \(Tc_{mix} = (Peercentage(CH_4) * Tc(CH_4)) + (Peercentage(C_2H_6) * Tc(C_2H_6)) + (Peercentage(C_2H_4) * Tc(C_2H_4)) \)
04

Calculation of Reduced Temperature and Pressure

Calculate the reduced temperature and pressure using the formula: \(Tr = T/Tc_{mix}\) and \(Pr = P/Pc_{mix}\) where actual gas temperature T= 90 + 273 = 363K and actual pressure P= 200bar. Note: Convert the temperature to Kelvins by adding 273 to Celsius.
05

Calculate Final Volume

Use the ideal gas law to calculate the final volume \(V= nRT/P\), where R= Ideal gas constant = 83.14(dimension depending on the units of P and V) and n = total moles = mass /molecular weight. The molecular weight is a weighted average calculated by multiplying the molecular weights of the individual gases by their percentages and summing the results: \(MW_{mix} = percentage(CH_4)*MW(CH_4) + percentage(C_2H_6)*MW(C_2H_6) +percentage(C_2H_4)*MW(C_2H_4)\), where \(MW(CH_4) = 16 kg/kmol, MW(C_2H_6) = 30 kg/kmol, MW(C_2H_4) = 28 kg/kmol\).

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Key Concepts

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

Mole Fractions
Understanding mole fractions is essential in mixture calculations. It's a way to express the composition of a mix without worrying about the actual amount of substance. A mole fraction is the ratio of the number of moles of a component to the total number of moles in the mixture. For example, if a mixture is composed of 20% methane, 30% ethane, and 50% ethylene by moles, we could say the mole fraction of methane is 0.2, of ethane is 0.3, and of ethylene is 0.5.

Why do we care about mole fractions? They allow us to make calculations on mixtures, like finding their average properties or behavior under certain conditions, without dealing with masses or volumes directly. This is especially helpful when we're working with gases, which can expand or compress significantly under different temperatures and pressures.
Pseudo Critical Pressure
The pseudo critical pressure concept is vital when dealing with gas mixtures like in Kay's rule for gas compression. It represents an average critical pressure for a mixture based on its composition. But what is critical pressure in the first place? It's the pressure needed to liquefy a gas at its critical temperature. The pseudo critical pressure is found by taking into account the mole fractions and critical pressures for each individual gas.

For our example with methane, ethane, and ethylene, calculating the pseudo critical pressure involves multiplying each component's mole fraction by its own critical pressure, then summing those numbers. This calculation doesn't just give us a number to plug into formulas—it provides insights into how the mixture behaves under compression compared to pure substances. Finding the intersection of such complex behavior requires us to understand and calculate these pseudo properties.
Pseudo Critical Temperature
Similar to pseudo critical pressure, the pseudo critical temperature is an average critical temperature for a gas mixture. The critical temperature itself is the highest temperature at which a substance can exist as a liquid, regardless of pressure. For mixtures, the pseudo critical temperature is crucial for understanding the thermodynamic behavior during processes like compression.

Using mole fractions and critical temperatures of each gas in our mixture, we find this pseudo value. It's done by multiplying the mole fraction of each gas by its critical temperature and then adding them up. This figure is necessary for finding reduced temperatures and pressures, which are used in various equations to predict a mixture's behavior under different conditions, especially when precise models like Kay's rule are applied.
Ideal Gas Law
The ideal gas law is a cornerstone in understanding the behavior of gases under various conditions. It's usually expressed as PV = nRT, where P stands for pressure, V for volume, n for moles of gas, R for the ideal gas constant, and T for temperature. This equation assumes that gas molecules don't interact with one another and occupy no space—though this isn't true for real gases, it's a useful approximation at standard conditions.

When we consider the compression of a gas mixture using Kay's rule, the ideal gas law can still apply under the assumption that the gas behaves ideally. We use it to calculate the final volume of the gas after compression, taking into account the total number of moles, the temperature in Kelvins, and the pressure in the appropriate units. It's an elegant expression revealing how closely interconnected those physical properties are.
Chemical Process Calculations
Performing chemical process calculations involves applying mathematical methods to chemical processes, allowing engineers and scientists to predict the outcomes under different scenarios. Variables like temperature, pressure, volume, and mole fractions are taken into account, often utilizing laws and rules like the ideal gas law and Kay's rule, respectively.

To reliably forecast the results of compressing our gas mixture, we must calculate the mixture's average molecular weight based on its components' molecular weights and their mole fractions. Then, with all these calculations—from determining mole fractions, pseudo critical properties, reduced temperatures and pressures, to finding average molecular weights—we're equipping ourselves to solve complex problems such as estimating final volumes of compressed gases. These skillsets and calculations are the backbone of designing and operating countless industrial processes in the chemical industry.

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Most popular questions from this chapter

A nitrogen rotameter is calibrated by feeding \(\mathrm{N}_{2}\) from a compressor through a pressure regulator, a needle valve, the rotameter, and a dry test meter, a device that measures the total volume of gas that passes through it. A water manometer is used to measure the gas pressure at the rotameter outlet. A flow rate is set using the needle valve, the rotameter reading, \(\phi\), is noted, and the change in the dry gas meter reading \((\Delta V)\) for a measured running time \((\Delta t)\) is recorded. The following calibration data are taken on a day when the temperature is \(23^{\circ} \mathrm{C}\) and barometric pressure is \(763 \mathrm{mm} \mathrm{Hg} .\) $$\begin{array}{rrr} \hline \phi & \Delta t(\min ) & \Delta V(\mathrm{L}) \\ \hline 5.0 & 10.0 & 1.50 \\ 9.0 & 10.0 & 2.90 \\ 12.0 & 5.0 & 2.00 \\ \hline \end{array}$$ (a) Prepare a calibration chart of \(\phi\) versus \(\dot{V}_{\text {sid }}\), the flow rate in standard \(\mathrm{cm}^{3} / \mathrm{min}\) equivalent to the actual flow rate at the measurement conditions. (b) Suppose the rotameter-valve combination is to be used to set the flow rate to 0.010 mol \(\mathrm{N}_{2} / \mathrm{min}\). What rotameter reading must be maintained by adjusting the valve?

The oxidation of nitric oxide $$\mathrm{NO}+\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{NO}_{2}$$ takes place in an isothermal batch reactor. The reactor is charged with a mixture containing 20.0 volume percent NO and the balance air at an initial pressure of \(380 \mathrm{kPa}\) (absolute). (a) Assuming ideal-gas behavior, determine the composition of the mixture (component mole fractions) and the final pressure (kPa) if the conversion of NO is 90\%. (b) Suppose the pressure in the reactor eventually equilibrates (levels out) at \(360 \mathrm{kPa}\). What is the equilibrium percent conversion of NO? Calculate the reaction equilibrium constant at the prevailing temperature, \(K_{p}\left[(\mathrm{atm})^{-0.5}\right]\), defined as $$K_{p}=\frac{\left(p_{\mathrm{NO}_{2}}\right)}{\left(p_{\mathrm{NO}}\right)\left(p_{\mathrm{O}_{2}}\right)^{0.5}}$$ where \(p_{i}(\mathrm{atm})\) is the partial pressure of species \(i\left(\mathrm{NO}_{2}, \mathrm{NO}, \mathrm{O}_{2}\right)\) at equilibrium. (c) Assuming that \(K_{\mathrm{p}}\) depends only on temperature, estimate the final pressure and composition in the reactor if the feed ratio of NO to \(\mathrm{O}_{2}\) and the initial pressure are the same as in \(\operatorname{Part}(\) a), but the feed to the reactor is pure \(\mathrm{O}_{2}\) instead of air. (d) Replace the partial pressures in the expression for \(K_{\mathrm{p}}\), and use the result to explain how reactor pressure influences the conversion of NO to \(\mathrm{NO}_{2}\)

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.)

The bacteria acetobacter aceti convert ethanol to acetic acid in the presence of oxygen according to the reaction $$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}+\mathrm{O}_{2} \rightarrow \mathrm{CH}_{3} \mathrm{COOH}+\mathrm{H}_{2} \mathrm{O}$$ In a continuous fermentation process, ethanol enters the top of the fermenter at a rate of \(145 \mathrm{kg} / \mathrm{h}\), and the air fed to the bottom of the fermenter is \(25 \%\) in excess of the amount required to consume all of the ethanol. A gas stream containing nitrogen and unreacted oxygen leaves the top of the fermenter, and a liquid stream containing acetic acid, water, and \(10 \%\) of the entering ethanol leaves the bottom. Assume that none of the ethanol, water, and acetic acid in the reactor is vaporized. The fermenter operates at \(30^{\circ} \mathrm{C},\) maintains a liquid \((\mathrm{SG}=0.95)\) height of \(4.5 \mathrm{m},\) and is open to the atmosphere (i.e., the pressure at the top of the fermenter is 1 atm). (a) What is the volumetric flow rate of air as it enters the bottom of the fermenter? What is the volumetric flow rate of gas leaving the top of the fermenter? (b) Assume a linear relationship between the fraction of oxygen reacted and the position of gas bubbles rising through the liquid in the fermenter: for example, half of the oxygen reacted is consumed in the bottom half of the fermenter. At the vertical midpoint of the fermenter, the average bubble diameter is \(1.5 \mathrm{mm}\). What is the average bubble diameter at the entry point of the air and as the gas leaves the liquid at the top of the fermenter?

A stream of oxygen enters a compressor at \(298 \mathrm{K}\) and 1.00 atm at a rate of \(127 \mathrm{m}^{3} / \mathrm{h}\) and is compressed to \(358 \mathrm{K}\) and 1000 atm. Estimate the volumetric flow rate of compressed \(\mathrm{O}_{2},\) using the compressibility-factor equation of state.

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