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Chapter 1: Problem 14

Calculate the mass of a mole of dry air, which is a mixture of \(\mathrm{N}_{2}\) \((78 \% \text { by volume }), \mathrm{O}_{2}(21 \%),\) and argon \((1 \%)\).

Short Answer

Expert verified
The mass of a mole of dry air is approximately 28.98 g/mol.

Step by step solution

01

Determine Molar Masses

To calculate the mass of a mole of dry air, we start by finding the molar masses of each gas in the mixture. - Nitrogen ( N_2): 28.02 g/mol - Oxygen ( O_2): 32.00 g/mol - Argon (Ar): 39.95 g/mol
02

Convert Volume Percentages to Mole Fractions

Since the gases are mixed by volume, we can use the volume percentages as mole percentages for our calculation. - Mole fraction of N_2: 0.78 - Mole fraction of O_2: 0.21 - Mole fraction of Ar: 0.01
03

Calculate Molar Mass of Dry Air

The molar mass of dry air is the sum of the products of the molar mass and mole fraction of each component:\[\text{Molar Mass of Dry Air} = (28.02 \, \text{g/mol} \times 0.78) + (32.00 \, \text{g/mol} \times 0.21) + (39.95 \, \text{g/mol} \times 0.01)\]Calculate each term:- N_2: 21.86 g/mol- O_2: 6.72 g/mol- Ar: 0.40 g/molAdd them together to get the total molar mass of dry air.
04

Summarize the Calculation

Add up the contributions from each gas to find the total molar mass:\[\text{Molar Mass of Dry Air} = 21.86 + 6.72 + 0.40 = 28.98 \, \text{g/mol}\]Thus, the mass of a mole of dry air is approximately 28.98 g/mol.

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Key Concepts

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

Mole Fraction
The concept of "mole fraction" is frequently used in chemistry to describe the composition of mixtures. It represents the ratio of the number of moles of a particular component to the total number of moles in the mixture. In simple terms, it tells us how much of one substance is present compared to the whole mixture.

To find the mole fraction, we use the following formula:
  • Mole Fraction (\(x_i\)) = Number of Moles of Component (\(n_i\)) / Total Number of Moles (\(n_{total}\))
For gases, the mole fraction is often equivalent to the volume percentage, since gases behave ideally under certain conditions. In the dry air example, nitrogen (\(\text{N}_2\)) has a mole fraction of 0.78, which means 78% of the air consists of nitrogen molecules. Similarly, oxygen (\(\text{O}_2\)) has a mole fraction of 0.21, and argon (\(\text{Ar}\)) has 0.01. The mole fractions are essential for calculations, especially when determining the properties of a mixture like molar mass.
Volume Percentage
Volume percentage is a way to express the concentration of a component in a gas mixture. It tells us how much of the volume of the mixture is taken up by a specific gas. This measure is particularly handy for gas mixtures because their components often combine volumetrically.

Since gases at the same temperature and pressure occupy equal volumes, volume percentages can directly translate into mole percentages. Therefore, in dry air, if nitrogen is present at 78% by volume, this means that 78% of the total volume of air is made up of nitrogen gas. The same idea applies to oxygen, at 21% by volume, and argon, at 1% by volume. This approach simplifies calculations by allowing us to use these percentages as mole fractions in further calculations.
Mixture Composition
Mixture composition in chemistry refers to how the different substances in a mixture are combined. For gases, the composition is often specified as parts of the whole, either by volume or moles, due to their ideal behavior under a range of conditions.

The mixture composition affects physical properties like density and molar mass. With dry air, composed of nitrogen, oxygen, and argon, we determine the molar mass by giving each component a weight according to its proportion, either in terms of volume percentages or mole fractions. Thus, knowing the composition allows chemists to predict how the mixture will behave and interact, which is vital in applications ranging from environmental science to industrial chemistry.
Gas Components
Understanding gas components means recognizing each individual element or compound present in a gaseous mixture. Each gas has its own properties, like molar mass, that contribute to the overall characteristics and behavior of the mixture. Nitrogen, oxygen, and argon in dry air are three primary components with distinct properties.

  • Nitrogen (\(\text{N}_2\)): The most abundant, known for its inertness and lightness.
  • Oxygen (\(\text{O}_2\)): Essential for combustion and respiration, contributing to a higher molar mass.
  • Argon (\(\text{Ar}\)): A noble gas that is chemically inactive, but adds to the molar mass due to its heavier atomic weight.
Each gas component affects the mixture鈥檚 properties, like the molar mass of dry air which we calculated to be 28.98 g/mol. This value helps in various scientific and practical calculations, such as determining the air鈥檚 density or its buoyancy in weather balloons.

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Most popular questions from this chapter

If you poke a hole in a container full of gas, the gas will start leaking out. In this problem you will make a rough estimate of the rate at which gas escapes through a hole. (This process is called effusion, at least when the hole is sufficiently small.) (a) Consider a small portion (area \(=A\) ) of the inside wall of a container full of gas. Show that the number of molecules colliding with this surface in a time interval \(\Delta t\) is \(P A \Delta t /(2 m \overline{v_{x}}),\) where \(P\) is the pressure, \(m\) is the average molecular mass, and \(\overline{v_{x}}\) is the average \(x\) velocity of those molecules that collide with the wall. (b) It's not easy to calculate \(\overline{v_{x}},\) but a good enough approximation is \((\overline{v_{x}^{2}})^{1 / 2}\) where the bar now represents an average over all molecules in the gas. Show that \((\overline{v_{x}^{2}})^{1 / 2}=\sqrt{k T / m}\) (c) If we now take away this small part of the wall of the container, the molecules that would have collided with it will instead escape through the hole. Assuming that nothing enters through the hole, show that the number \(N\) of molecules inside the container as a function of time is governed by the differential equation $$\frac{d N}{d t}=-\frac{A}{2 V} \sqrt{\frac{k T}{m}} N$$ Solve this equation (assuming constant temperature) to obtain a formula of the form \(N(t)=N(0) e^{-t / \tau},\) where \(\tau\) is the "characteristic time" for \(N\) \((\text { and } P)\) to drop by a factor of \(e\) (d) Calculate the characteristic time for a gas to escape from a 1 -liter container punctured by a \(1-\mathrm{mm}^{2}\) hole. (e) Your bicycle tire has a slow leak, so that it goes flat within about an hour after being inflated. Roughly how big is the hole? (Use any reasonable estimate for the volume of the tire.) (f) In Jules Verne's Round the Moon, the space travelers dispose of a dog's corpse by quickly opening a window, tossing it out, and closing the window. Do you think they can do this quickly enough to prevent a significant amount of air from escaping? Justify your answer with some rough estimates and calculations.

Estimate how long it should take to bring a cup of water to boiling temperature in a typical 600 -watt microwave oven, assuming that all the energy ends up in the water. (Assume any reasonable initial temperature for the water.) Explain why no heat is involved in this process.

When spring finally arrives in the mountains, the snow pack may be two meters deep, composed of \(50 \%\) ice and \(50 \%\) air. Direct sunlight provides about 1000 watts/m \(^{2}\) to earth's surface, but the snow might reflect \(90 \%\) of this energy. Estimate how many weeks the snow pack should last, if direct solar radiation is the only source of energy.

Imagine some helium in a cylinder with an initial volume of 1 liter and an initial pressure of 1 atm. Somehow the helium is made to expand to a final volume of 3 liters, in such a way that its pressure rises in direct proportion to its volume.

During a hailstorm, hailstones with an average mass of 2 g and a speed of \(15 \mathrm{m} / \mathrm{s}\) strike a window pane at a \(45^{\circ}\) angle. The area of the window is \(0.5 \mathrm{m}^{2}\) and the hailstones hit it at a rate of 30 per second. What average pressure do they exert on the window? How does this compare to the pressure of the atmosphere?
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