/*! 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 74 (a) What is the number of molecu... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(a) What is the number of molecules per cubic meter in air at \(20^{\circ} \mathrm{C}\) and at a pressure of \(1.0 \mathrm{~atm}\left(=1.01 \times 10^{5} \mathrm{~Pa}\right) ?(\mathrm{~b}) \mathrm{What}\) is the mass of \(1.0 \mathrm{~m}^{3}\) of this air? Assume that \(75 \%\) of the molecules are nitrogen \(\left(\mathrm{N}_{2}\right)\) and \(25 \%\) are oxygen \(\left(\mathrm{O}_{2}\right) .\)

Short Answer

Expert verified
Approximately \(2.50 \times 10^{25}\) molecules per cubic meter; mass of 1.0 m³ of air is about 1.206 kg.

Step by step solution

01

Identify the Known Values

We are given the temperature of the air as \(20^{\circ} \mathrm{C}\), which equals \(293.15\, \mathrm{K}\) when converted to Kelvin. We are also given the pressure as \(1.01 \times 10^5 \, \mathrm{Pa}\). Furthermore, we know that \(75\%\) of the molecules are nitrogen (\(\mathrm{N}_2\)) and \(25\%\) are oxygen (\(\mathrm{O}_2\)).
02

Use the Ideal Gas Law to Find Number Density

The ideal gas law is \( PV = nRT \), where \( n \) is the number of moles per cubic meter. Rearrange it to find \( n \):\[ n = \frac{P}{RT} \]With \( R = 8.314 \, \mathrm{J/(mol \cdot K)} \) (universal gas constant) and substituting \( P \) and \( T \), we calculate:\[ n = \frac{1.01 \times 10^5}{8.314 \times 293.15} \approx 41.6 \, \mathrm{mol/m^3} \]
03

Calculate the Number of Molecules per Cubic Meter

Use Avogadro's number (\(6.022 \times 10^{23}\, \mathrm{molecules/mol}\)) to find the number of molecules per cubic meter:\[ \text{Number of molecules} = n \times 6.022 \times 10^{23} \approx 41.6 \times 6.022 \times 10^{23} \approx 2.50 \times 10^{25} \]
04

Calculate Molar Mass of Air

The molar mass of nitrogen \(\mathrm{N}_2\) is \(28 \, \mathrm{g/mol}\) and oxygen \(\mathrm{O}_2\) is \(32 \, \mathrm{g/mol}\). Since \(75\%\) of the molecules are \(\mathrm{N}_2\) and \(25\%\) are \(\mathrm{O}_2\), the average molar mass is:\[ M = 0.75 \times 28 + 0.25 \times 32 = 29 \, \mathrm{g/mol} \]
05

Calculate Mass of 1.0 m³ of Air

Using the number of moles per cubic meter from Step 2 and the average molar mass, the mass of air is:\[ \text{Mass} = n \times M = 41.6 \, \mathrm{mol/m^3} \times 29 \, \mathrm{g/mol} \approx 1206.4 \, \mathrm{g/m^3} \]
06

Convert Mass to Kilograms

Convert the mass from grams to kilograms since the result should be in standard units for mass:\[ 1206.4 \, \mathrm{g/m^3} = 1.2064 \, \mathrm{kg/m^3} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Number of Molecules
When discussing gases, it is crucial to understand the number of molecules present in a given volume. To determine the number of molecules in air, we use the ideal gas law, which ties together pressure (\(P\)), volume (\(V\)), and temperature (\(T\)) with the number of moles (\(n\)) of the gas. The ideal gas equation is: \[PV = nRT\] where \(R\) is the universal gas constant.
To find the number of molecules in a cubic meter of air, we first rearrange the equation to solve for \(n\): \[n = \frac{P}{RT}\]. Given the pressure and temperature, we can compute \(n\), which represents the moles per cubic meter.
Once \(n\) is known, we multiply it by Avogadro's number (\(6.022 \times 10^{23}\, \text{molecules/mol}\)) to obtain the actual number of molecules per cubic meter: \[\text{Number of molecules} = n \times 6.022 \times 10^{23}\]. This formula helps us understand how densely packed the molecules are in a given volume of air.
Molar Mass
Molar mass is an important concept when discussing the composition of gases, such as air. It tells us the mass of one mole of a substance, in grams per mole (g/mol). To find the molar mass of air, which is a mixture of various gases, we must account for the molar masses of each component.
Air is composed primarily of nitrogen (\(\text{N}_2\)) and oxygen (\(\text{O}_2\)), making up approximately 75% and 25% of the volume, respectively. Knowing their molar masses (\(28\,\text{g/mol}\) for nitrogen and \(32\,\text{g/mol}\) for oxygen), we calculate the average molar mass as follows:
  • Multiply the molar mass of nitrogen by its proportion: \(0.75 \times 28\,\text{g/mol} = 21\,\text{g/mol}\).
  • Multiply the molar mass of oxygen by its proportion: \(0.25 \times 32\,\text{g/mol} = 8\,\text{g/mol}\).
  • Add these results together: \(21\,\text{g/mol} + 8\,\text{g/mol} = 29\,\text{g/mol}\).
Hence, the molar mass of air is approximately \(29\,\text{g/mol}\). Understanding this allows us to further compute the mass of air per cubic meter, facilitating studies related to atmospheric conditions and chemistry.
Avogadro's Number
Avogadro's number is a fundamental constant in chemistry that quantifies the number of entities (typically atoms or molecules) in one mole of a substance. This constant is approximately \(6.022 \times 10^{23}\) entities per mole.
Avogadro's number enables scientists to convert between the macroscopic scale of moles and the microscopic scale of individual molecules. For example, once we calculate the number of moles of gas in a specific volume using the ideal gas law, we can multiply that value by Avogadro's number to determine how many molecules are present.
In practical applications, such as determining the number of molecules in air, this constant links the measurable, bulk concentrations of gases to their molecular identities. Using Avogadro's number is crucial when dealing with gases at a molecular level, allowing for accurate scientific predictions and calculations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An ideal gas with \(3.00 \mathrm{~mol}\) is initially in state 1 with pressure \(p_{1}=20.0 \mathrm{~atm}\) and volume \(V_{1}=1500 \mathrm{~cm}^{3} .\) First it is taken to state 2 with pressure \(p_{2}=1.50 p_{1}\) and volume \(V_{2}=2.00 V_{1} .\) Then it is taken to state 3 with pressure \(p_{3}=2.00 p_{1}\) and volume \(V_{3}=0.500 V_{1} .\) What is the temperature of the gas in (a) state 1 and (b) state \(2 ?\) (c) What is the net change in internal energy from state 1 to state \(3 ?\)

The envelope and basket of a hot-air balloon have a combined weight of \(2.45 \mathrm{kN}\), and the envelope has a capacity (volume) of \(2.18 \times 10^{3} \mathrm{~m}^{3} .\) When it is fully inflated, what should be the temperature of the enclosed air to give the balloon a lifting capacity (force) of \(2.67 \mathrm{kN}\) (in addition to the balloon's weight)? Assume that the surrounding air, at \(20.0^{\circ} \mathrm{C},\) has a weight per unit volume of \(11.9 \mathrm{~N} / \mathrm{m}^{3}\) and a molecular mass of \(0.028 \mathrm{~kg} / \mathrm{mol},\) and is at a pressure of \(1.0 \mathrm{~atm} .\)

A container encloses \(2 \mathrm{~mol}\) of an ideal gas that has molar mass \(M_{1}\) and 0.5 mol of a second ideal gas that has molar mass \(M_{2}=3 M_{1}\). What fraction of the total pressure on the container wall is attributable to the second gas? (The kinetic theory explanation of pressure leads to the experimentally discovered law of partial pressures for a mixture of gases that do not react chemically: The total pressure exerted by the mixture is equal to the sum of the pressures that the several gases would exert separately if each were to occupy the vessel alone. The molecule-vessel collisions of one type would not be altered by the presence of another type.)

A quantity of ideal gas at \(10.0^{\circ} \mathrm{C}\) and \(100 \mathrm{kPa}\) occupies a volume of \(2.50 \mathrm{~m}^{3}\). (a) How many moles of the gas are present? (b) If the pressure is now raised to \(300 \mathrm{kPa}\) and the temperature is raised to \(30.0^{\circ} \mathrm{C}\), how much volume does the gas occupy? Assume no leaks.

What is the average translational kinetic energy of nitrogen molecules at \(1600 \mathrm{~K} ?\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.