Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 37

Air at room temperature and atmospheric pressure has a mass density of $1.2 \mathrm{kg} / \mathrm{m}^{3} .$ The average molecular mass of air is 29.0 u. How many molecules are in \(1.0 \mathrm{cm}^{3}\) of air?

Short Answer

Expert verified
Answer: Approximately \(2.49 \times 10^{19}\) air molecules are present in 1.0 cm³ of air.

Step by step solution

01

Convert mass density of air

First, we need to convert the mass density of air from kg/m³ to g/cm³. We know that 1 kg = 1000 g and 1 m = 100 cm. Therefore, \(1.2 \: \text{kg/m}^3 \times \frac{1000 \: \text{g}}{1 \: \text{kg}} \times \left(\frac{1 \: \text{m}}{100 \: \text{cm}}\right)^3 = 1.2 \times 10^{-3} \: \text{g/cm}^3\)
02

Calculate total mass in 1 cm³

Now, let's find the total mass of 1 cm³ of air by multiplying the mass density by the volume: \(1.2 \times 10^{-3} \: \text{g/cm}^3 \times 1.0 \: \text{cm}^3 = 1.2 \times 10^{-3} \: \text{g}\)
03

Calculate moles of air molecules

Since the average molecular mass of air is given as 29.0 u, we can use this to calculate the moles of air molecules in 1 cm³ of air. To do this, we need to divide the total mass of air molecules by the average molecular mass: \(\text{Moles of air molecules} = \frac{\text{Total mass of air molecules}}{\text{Average molecular mass of air}}\) \(\text{Moles of air molecules} = \frac{1.2 \times 10^{-3} \: \text{g}}{29.0 \: \text{g/mol}} = 4.14 \times 10^{-5} \: \text{mol}\)
04

Use Avogadro's number to find the number of air molecules

Finally, we can use Avogadro's number (\(6.022 \times 10^{23} \: \text{molecules/mol}\)) to convert the moles of air molecules into the actual number of molecules: \(\text{Number of air molecules} = \text{Moles of air molecules} \times \text{Avogadro's number}\) \(\text{Number of air molecules} = 4.14 \times 10^{-5} \: \text{mol} \times 6.022 \times 10^{23} \: \text{molecules/mol} \approx 2.49 \times 10^{19} \: \text{molecules}\) So, there are approximately \(2.49 \times 10^{19}\) air molecules in 1.0 cm³ of air.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

A temperature change \(\Delta T\) causes a volume change \(\Delta V\) but has no effect on the mass of an object. (a) Show that the change in density $\Delta \rho\( is given by \)\Delta \rho=-\beta \rho \Delta T$. (b) Find the fractional change in density \((\Delta \rho / \rho)\) of a brass sphere when the temperature changes from \(32^{\circ} \mathrm{C}\) to \(-10.0^{\circ} \mathrm{C}\)
As a Boeing 747 gains altitude, the passenger cabin is pressurized. However, the cabin is not pressurized fully to atmospheric $\left(1.01 \times 10^{5} \mathrm{Pa}\right),$ as it would be at sea level, but rather pressurized to \(7.62 \times 10^{4} \mathrm{Pa}\). Suppose a 747 takes off from sea level when the temperature in the airplane is \(25.0^{\circ} \mathrm{C}\) and the pressure is \(1.01 \times 10^{5} \mathrm{Pa} .\) (a) If the cabin temperature remains at \(25.0^{\circ} \mathrm{C},\) what is the percentage change in the number of moles of air in the cabin? (b) If instead, the number of moles of air in the cabin does not change, what would the temperature be?
A flat square of side \(s_{0}\) at temperature \(T_{0}\) expands by \(\Delta s\) in both length and width when the temperature increases by \(\Delta T\). The original area is \(s_{0}^{2}=A_{0}\) and the final area is $\left(s_{0}+\Delta s\right)^{2}=A .\( Show that if \)\Delta s \ll s_{0}$$$\frac{\Delta A}{A_{0}}=2 \alpha \Delta T$$(Although we derive this relation for a square plate, it applies to a flat area of any shape.)
A hydrogen balloon at Earth's surface has a volume of \(5.0 \mathrm{m}^{3}\) on a day when the temperature is \(27^{\circ} \mathrm{C}\) and the pressure is \(1.00 \times 10^{5} \mathrm{N} / \mathrm{m}^{2} .\) The balloon rises and expands as the pressure drops. What would the volume of the same number of moles of hydrogen be at an altitude of \(40 \mathrm{km}\) where the pressure is \(0.33 \times 10^{3} \mathrm{N} / \mathrm{m}^{2}\) and the temperature is \(-13^{\circ} \mathrm{C} ?\)
The mass of \(1 \mathrm{mol}\) of \(^{13} \mathrm{C}\) (carbon- 13 ) is $13.003 \mathrm{g}$. (a) What is the mass in \(u\) of one \(^{13} \mathrm{C}\) atom? (b) What is the mass in kilograms of one \(^{13} \mathrm{C}\) atom?
See all solutions

Recommended explanations on Physics Textbooks

Work Energy and Power

Read Explanation

Kinematics Physics

Read Explanation

Modern Physics

Read Explanation

Solid State Physics

Read Explanation

Scientific Method Physics

Read Explanation

Force

Read Explanation
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.

Sign-up for free