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Chapter 5: Problem 34

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

Lewis \(^{12}\) describes the hazards of breathing air containing appreciable amounts of an asphyxiant (a gas that has no specific toxicity but, when inhaled, excludes oxygen from the lungs). When the mole percent of the asphyxiant in the air reaches \(50 \%,\) marked symptoms of distress appear, and at \(75 \%\) death occurs in a matter of minutes. A small storage room whose dimensions are \(2 \mathrm{m} \times 1.5 \mathrm{m} \times 3 \mathrm{m}\) contains a number of expensive and dangerous chemicals. To prevent unauthorized entry, the door to the room is always locked and can be opened with a key from either side. A cylinder of liquid carbon dioxide is stored in the room. The valve on the cylinder is faulty and some of the contents have escaped over the weekend. The room temperature is \(25^{\circ} \mathrm{C}\). (a) If the concentration of \(\mathrm{CO}_{2}\) reaches the lethal 75 mole \(\%\) level, what would be the mole percent of \(\mathrm{O}_{2} ?\) (b) How much \(\mathrm{CO}_{2}(\mathrm{kg})\) is present in the room when the lethal concentration is reached? Why would more than that amount have to escape from the cylinder for this concentration to be reached? (c) Describe a set of events that could result in a fatality in the given situation. Suggest at least two measures that would reduce the hazards associated with storage of this scemingly harmless substance.

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