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An ideal-gas mixture contains \(35 \%\) helium, \(20 \%\) methane, and \(45 \%\) nitrogen by volume at 2.00 atm absolute and \(90^{\circ} \mathrm{C}\). Calculate (a) the partial pressure of each component, (b) the mass fraction of methane, (c) the average molecular weight of the gas, and (d) the density of the gas in \(\mathrm{kg} / \mathrm{m}^{3}\).

Short Answer

Expert verified
The results are: (a) Partial pressures: Helium: 0.70 atm, Methane: 0.40 atm, Nitrogen: 0.90 atm, (b) Mass fraction of Methane: 0.19, (c) Average molecular weight: 17.04 g/mol, (d) Density of gas: 1.38 kg/m^3.

Step by step solution

01

Calculation of partial pressures

The partial pressure of a gas is given by the total pressure times the volume fraction of that gas. Thus, the partial pressures will be: - For helium: \(0.35 脳 2.00 atm = 0.70 atm\), - For methane: \(0.20 脳 2.00 atm = 0.40 atm\), - For nitrogen: \(0.45 脳 2.00 atm = 0.90 atm\).
02

Mass fraction calculation for methane

The mass of each component is given by the partial percentage multiplied by its molecular weight, and then the mass fraction is found by dividing the mass by the total mass. The molar masses are: Helium (He) = 4 \(g/mol\), Methane (CH4) = 16 \(g/mol\), Nitrogen (N2) = 28 \(g/mol\). Therefore, the mass of each component will be: - For He: \(0.35 脳 4 = 1.4 g\), - For CH4: \(0.20 脳 16 = 3.2 g\), - For N2: \(0.45 脳 28 = 12.6 g\). The sum of these gives the total mass, which is \(1.4 g + 3.2 g + 12.6 g = 17.2 g\). The mass fraction of methane is then the mass of methane divided by the total mass, which is \(3.2 g/17.2 g = 0.19\).
03

Average molecular weight calculation

The average molecular weight is found by weighting the molecular mass by the mass fraction of each component. This is given by: Average molecular weight = \(0.35 脳 4 + 0.19 脳 16 + 0.45 脳 28 = 1.4 g/mol + 3.04 g/mol +12.6 g/mol = 17.04 g/mol\).
04

Calculation of gas density

For this, we will use the equation of state for ideal gases: \(P = 蟻RT/M\), where P is the total pressure, 蟻 is the density, R is the universal gas constant = 0.0821 \(atm.L/mol.K\), T is the temperature in Kelvins and M is the average molar mass in kg per mole. First, convert the temperature to Kelvins: \(90^{\circ}C = 90+273.15 = 363.15 K\). The density is given by 蟻 = \(PM/RT = 2.00 atm 脳 17.04 脳 10^{-3} kg/mol / (0.0821 atm.L/mol.K 脳 363.15 K) = 1.38 kg/m^3\).

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Key Concepts

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

Partial Pressure Calculation
Understanding the concept of partial pressure in an ideal gas mixture is important. Each gas in a mixture contributes to the total pressure based on its volume percentage.

The formula used to calculate the partial pressure of each component is simple. It's a product of the total pressure and the volume fraction of that gas.
  • For helium: Given volume fraction is 35%. So, its partial pressure is calculated as \(0.35 \times 2.00 \text{ atm} = 0.70 \text{ atm}\).
  • Methane, with a 20% volume fraction, has a partial pressure of \(0.20 \times 2.00 \text{ atm} = 0.40 \text{ atm}\).
  • Nitrogen holds 45% of the volume, leading to a partial pressure of \(0.45 \times 2.00 \text{ atm} = 0.90 \text{ atm}\).
Partial pressures are crucial as they directly affect reactions and dynamics in the gas mixture.
Mass Fraction Calculation
Mass fraction helps to measure the contribution of a single gas component in terms of mass within a mixture.

Calculating mass fractions involves a few steps, using molecular weights and given percentages.
  • The molar mass of helium is 4 \(\text{g/mol}\), methane is 16 \(\text{g/mol}\), and nitrogen is 28 \(\text{g/mol}\).
  • You first determine the mass of each gas:
    • Helium: \(0.35 \times 4 = 1.4 \text{g}\)
    • Methane: \(0.20 \times 16 = 3.2 \text{g}\)
    • Nitrogen: \(0.45 \times 28 = 12.6 \text{g}\)
  • Sum of these masses provides the total mass of the mixture: \(1.4 + 3.2 + 12.6 = 17.2 \text{g}\).
  • The mass fraction of methane is calculated by dividing its mass by the total mass: \(3.2 \text{g} / 17.2 \text{g} = 0.19\).
This fraction illustrates the weight contribution each gas partakes in the mixture.
Average Molecular Weight
The average molecular weight tells us about the typical mass of a molecule in a gas mixture. It's derived by considering both the mass fractions and the molecular weights.

The calculation mixes together the contribution of each gas component.
  • Use mass fractions: Helium contributes \(0.35 \), Methane \(0.19\), and Nitrogen \(0.45\).
  • Calculate average molecular weight:
    • Helium's contribution: \(0.35 \times 4 = 1.4 \text{ g/mol}\)
    • Methane's contribution: \(0.19 \times 16 = 3.04 \text{ g/mol}\)
    • Nitrogen's contribution: \(0.45 \times 28 = 12.6 \text{ g/mol}\)
  • The sum gives an average molecular weight of the mixture: \(1.4 + 3.04 + 12.6 = 17.04 \text{ g/mol}\).
This average molecular weight is vital for understanding the gas behavior under various conditions.
Gas Density Calculation
Calculating the density of a gas involves understanding the equation of state for ideal gases. It's influenced by pressure, temperature, and the gas's molecular weight.

Here's how you compute the density:
  • The equation used is \( P = \rho \frac{RT}{M} \), where:
    • \( P \) is total pressure.
    • \( \rho \) is density.
    • \( R \) is the universal gas constant \(0.0821 \text{ atm.L/mol.K}\).
    • \( T \) is the temperature in Kelvin. Convert Celsius to Kelvin: \(90^{\circ}C = 363.15 \text{ K}\).
    • \( M \) is the average molar mass \(17.04 \text{ g/mol} = 17.04 \times 10^{-3} \text{ kg/mol}\).
  • Rearrange the formula to find density: \[ \rho = \frac{PM}{RT} \]
  • Plug in the values to get: \( \rho = \frac{2.00 \times 17.04 \times 10^{-3}}{0.0821 \times 363.15} = 1.38 \text{ kg/m}^3\).
This final density value tells us about the mass of the gas per unit volume.

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Most popular questions from this chapter

An adult takes about 12 breaths per minute, inhaling roughly \(500 \mathrm{mL}\) of air with each breath. The molar compositions of the inspired and expired gases are as follows: $$\begin{array}{lcc} \hline \text { Species } & \text { Inspired Gas (\%) } & \text { Expired Gas (\%) } \\ \hline \mathrm{O}_{2} & 20.6 & 15.1 \\ \mathrm{CO}_{2} & 0.0 & 3.7 \\ \mathrm{N}_{2} & 77.4 & 75.0 \\ \mathrm{H}_{2} \mathrm{O} & 2.0 & 6.2 \\ \hline \end{array}$$ The inspired gas is at \(24^{\circ} \mathrm{C}\) and 1 atm, and the expired gas is at body temperature and pressure \(\left(37^{\circ} \mathrm{C}\right.\) and 1 atm). Nitrogen is not transported into or out of the blood in the lungs, so that \(\left(\mathrm{N}_{2}\right)_{\text {in }}=\left(\mathrm{N}_{2}\right)_{\text {out }}\) (a) Calculate the masses of \(\mathrm{O}_{2}, \mathrm{CO}_{2},\) and \(\mathrm{H}_{2} \mathrm{O}\) transferred from the pulmonary gases to the blood or vice versa (specify which) per minute. (b) Calculate the volume of air exhaled per milliliter inhaled. (c) At what rate (g/min) is this individual losing weight by merely breathing? (d) The rate at which oxygen is transferred from the air in the lungs to the blood is roughly proportional to \(\left[\left(p_{\mathrm{O}_{2}}\right)_{\mathrm{air}}-\left(p_{\mathrm{O}_{2}}\right)_{\mathrm{blood}}\right],\) where \(\left(p_{\mathrm{O}_{2}}\right)_{\mathrm{blood}}\) is a quantity related to the concentration of oxygen in the blood. Compared to regions where atmospheric pressure is 14.7 psia, what effect does the atmospheric pressure in Denver, which is approximately 12.1 psi, have on the transport rate and breathing rate? How does the body adjust to address this condition?

A stream of liquid \(n\) -pentane flows at a rate of \(50.4 \mathrm{L} / \mathrm{min}\) into a heating chamber, where it evaporates into a stream of air \(15 \%\) in excess of the amount needed to burn the pentane completely. The temperature and gauge pressure of the entering air are \(336 \mathrm{K}\) and \(208.6 \mathrm{kPa}\). The pentane-laden heated gas flows into a combustion furnace in which a fraction of the pentane is burned. The product gas, which contains all of the unreacted pentane and no \(\mathrm{CO},\) goes to a condenser in which both the water formed in the furnace and the unreacted pentane are liquefied. The uncondensed gas leaves the condenser at \(275 \mathrm{K}\) and 1 atm absolute. The liquid condensate is separated into its components, and the flow rate of the pentane is measured and found to be \(3.175 \mathrm{kg} / \mathrm{min}\). (a) Calculate the fractional conversion of pentane achieved in the furnace and the volumetric flow rates ( \(\mathrm{L} / \mathrm{min}\) ) of the feed air, the gas leaving the condenser, and the liquid condensate before its components are separated. (b) Sketch the apparatus that could have been used to separate the pentane and water in the condensate. Hint: Remember that pentane is a hydrocarbon and recall what is said about oil (hydrocarbons) and water.

In froth flotation, air is bubbled through an aqueous solution or slurry to which a foaming agent (soap) has been added. The air-soap bubbles carry finely dispersed solids and hydrophobic materials such as grease and oil to the surface where they can be skimmed off in the foam. An ore-containing slurry is to be processed in a froth flotation tank at a rate of 300 tons/h. The slurry consists of \(20.0 \mathrm{wt} \%\) solids (the ore, \(\mathrm{SG}=1.2\) ) and the remainder an aqueous solution with a density close to that of water. Air is sparged (blown through a nozzle designed to produce small bubbles) into the slurry at a rate of \(40.0 \mathrm{ft}^{3}\) (STP)/1000 gal of slurry. The entry point of the air is 10 \(\mathrm{ft}\) below the slurry surface. The tank contents are at \(75^{\circ} \mathrm{F}\) and the barometric pressure is 28.3 inches of Hg. The sparger design is such that the average bubble diameter on entry is \(2.0 \mathrm{mm}\). (a) What is the volumetric flow rate of the air at its entering conditions? (b) By what percentage does the average bubble diameter change between the entry point and the slurry surface?

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.

A certain gas has a molecular weight of \(30.0,\) a critical temperature of \(310 \mathrm{K}\), and a critical pressure of 4.5 MPa. Calculate the density in \(\mathrm{kg} / \mathrm{m}^{3}\) of this gas at \(465 \mathrm{K}\) and \(9.0 \mathrm{MPa}\) (a) if the gas is ideal and (b) if the gas obeys the law of corresponding states.

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