(a) If the pressure exerted by ozone, \(\mathrm{O}_{3},\) in the stratosphere is \(3.0 \times 10^{-3}\) atm and the temperature is 250 \(\mathrm{K}\) , how many ozone molecules are in a liter? (b) Carbon dioxide makes up approximately 0.04\(\%\) of Earth's atmosphere. If you collect a \(2.0-\mathrm{L}\) sample from the atmosphere at sea level \((1.00 \mathrm{atm})\) on a warm day \(\left(27^{\circ} \mathrm{C}\right),\) how many \(\mathrm{CO}_{2}\) molecules are in your sample?

Short Answer

Expert verified
(a) There are approximately \(1.81 \times 10^{18}\) ozone molecules in a liter. (b) There are approximately \(1.95 \times 10^{16}\) CO₂ molecules in the 2.0-L sample.

Step by step solution

01

Convert the given conditions to appropriate units

: We have the pressure in atm and the volume is given as 1 liter, which is equivalent to 0.001 m^3. Temperature is given in Kelvin, which is already in the required unit for the Ideal Gas Law.
02

Use the Ideal Gas Law to find moles

: We are given pressure P = \(3.0 \times 10^{-3}\) atm, temperature T = 250 K, and volume V = 0.001 m³. We will use the Ideal Gas Law: PV = nRT And we will use the gas constant value for R = 0.0821 L.atm/mol.K. We need to find the number of moles, n: n = PV/(RT)
03

Calculate the number of moles

: n = \((3.0 \times 10^{-3})(0.001)\) / \((0.0821)(250)\) n = \(3.0 \times 10^{-6}\) moles
04

Calculate the number of molecules

: Using Avogadro's number, which is 6.022 x \(10^{23}\) atoms/mol, we can find the number of molecules: Number of molecules = n (Number of ozone molecules per mole) Number of ozone molecules = \(3.0 \times 10^{-6}\) moles × 6.022 x \(10^{23}\) ozone molecules/mol Number of ozone molecules ≈ \(1.81 \times 10^{18}\) ozone molecules. Answer for Part (a): There are approximately \(1.81 \times 10^{18}\) ozone molecules in a liter. (b) Number of CO₂ molecules in 2.0-L sample at given pressure, temperature, and percentage in Earth's atmosphere
05

Convert the given conditions to appropriate units

: We have pressure (1.00 atm), temperature (27 °C, which is 300 K), and volume (2.0 L, which is 0.002 m³) given in proper units.
06

Calculate the moles of air in the sample

: We will use the Ideal Gas Law to find the moles of air in the given sample: n = PV/(RT) n = (1.00 × 0.002) / (0.0821 × 300) n ≈ 8.1 × \(10^{-5}\) moles of air
07

Calculate the moles of CO₂ in the sample

: Carbon dioxide makes up approximately 0.04% of Earth's atmosphere. Therefore, to find the moles of CO₂ in the sample, we multiply the moles of air by 0.0004 (0.04% in decimal form): Moles of CO₂ = 8.1 × \(10^{-5}\) × 0.0004 Moles of CO₂ ≈ 3.24 × \(10^{-8}\) moles
08

Calculate the number of CO₂ molecules

: To find the number of CO₂ molecules in the sample, we multiply the moles of CO₂ by Avogadro's number: Number of CO₂ molecules = 3.24 × \(10^{-8}\) moles × 6.022 x \(10^{23}\) molecules/mol Number of CO₂ molecules ≈ \(1.95 \times 10^{16}\) CO₂ molecules. Answer for Part (b): There are approximately \(1.95 \times 10^{16}\) CO₂ molecules in the 2.0-L sample.

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 Vaia!

Key Concepts

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

Avogadro's Number
One of the fundamental constants in chemistry and physics is Avogadro's number, which is approximately 6.022 x 1023. This staggering number represents the quantity of atoms, ions, or molecules in one mole of substance. Understanding this concept is key when dealing with the molecular scale.

When we discuss gases at the molecular level, Avogadro's number helps us convert from moles, which is a chemical unit of measurement for amount of substance, to actual numbers of particles. This conversion is essential in exercises involving the Ideal Gas Law, as it provides the link between an amount of gas in moles and the total number of molecules present.

In the given exercise, after calculating the number of moles of ozone using the Ideal Gas Law, it was necessary to multiply by Avogadro's number to determine the total molecules. This illustrates the importance of Avogadro's number in making sense of molecular quantities for real-world applications, such as measuring ozone concentration in the atmosphere.
Atmospheric Composition
The Earth's atmosphere is composed of a complex mixture of different gases. Nitrogen and oxygen are the most abundant, accounting for about 78% and 21% of the atmosphere by volume, respectively. The remaining 1% contains trace gases like argon, carbon dioxide (CO2), neon, and ozone (O3), each contributing to our planet’s climate and life support systems in unique ways.

Understanding the composition of the atmosphere is crucial for exercises involving the Ideal Gas Law. For example, when calculating the amount of a specific gas in a mixture, such as CO2 in the exercise, we must take into account its percentage composition. In the exercise, CO2 was noted to make up roughly 0.04% of the Earth's atmosphere. Using that percentage, one can calculate the moles of CO2 in a given volume of air and follow up with the calculation of molecules, as shown by multiplying it with Avogadro's number.
Gas Constant
The gas constant, often denoted by the symbol R, is an intrinsic value used in gas-related calculations. Specifically, its value is utilized in the equation stated by the Ideal Gas Law: PV = nRT. The gas constant bridges pressure, temperature, and volume with the amount of substance in moles.

In the step-by-step solution of the textbook exercise, R is given a value of 0.0821 L.atm/mol.K. This particular value of R is suited for pressure in atmospheres, volume in liters, and temperature in Kelvin. In different contexts, such as when working with energy, R may have different units like Joules per mole per Kelvin (J/mol·K).

The gas constant is universal and allows us to apply the Ideal Gas Law across various conditions and materials. By using R, we can solve for any one of the variables in the Ideal Gas Law, provided we have the other three. The gas constant is therefore an essential tool in our exercise, enabling the calculation of both ozone and CO2 molecules in the atmosphere when given pressure, volume, and temperature.

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 gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required 105 s for 1.0 L of the gas to effuse. Under identical experimental conditions it required 31 s for 1.0 \(\mathrm{L}\) of \(\mathrm{O}_{2}\) gas to effuse. Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of \(1.0 \mathrm{L} ;\) in other words, rate is the amount that diffuses over the time it takes to diffuse.)

An herbicide is found to contain only \(\mathrm{C}, \mathrm{H}, \mathrm{N},\) and Cl. The complete combustion of a 100.0 -mg sample of the herbicide in excess oxygen produces 83.16 \(\mathrm{mL}\) of \(\mathrm{CO}_{2}\) and 73.30 \(\mathrm{mL}\) of \(\mathrm{H}_{2} \mathrm{O}\) vapor expressed at STP. A separate analysis shows that the sample also contains 16.44 \(\mathrm{mg}\) of Cl. (a) Determine the percentage of the composition of the substance. ( b) Calculate its empirical formula. (c) What other information would you need to know about this compound to calculate its true molecular formula?

Cyclopropane, a gas used with oxygen as a general anesthetic, is composed of 85.7\(\% \mathrm{C}\) and 14.3\(\% \mathrm{H}\) by mass. (a) If 1.56 \(\mathrm{g}\) of cyclopropane has a volume of 1.00 \(\mathrm{L}\) at 0.984 atm and \(50.0^{\circ} \mathrm{C},\) what is the molecular formula of cyclopropane? (b) Judging from its molecular formula, would you expect cyclopropane to deviate more or less than Ar from ideal-gas behavior at moderately high pressures and room temperature? Explain. (c) Would cyclopropane effuse through a pinhole faster or more slowly than methane, CH. \(_{4} ?\)

A fixed quantity of gas at \(21^{\circ} \mathrm{C}\) exhibits a pressure of 752 torr and occupies a volume of 5.12 L. \((\mathbf{a})\) Calculate the volume the gas will occupy if the pressure is increased to 1.88 atm while the temperature is held constant. \((\mathbf{b})\) Calculate the volume the gas will occupy if the temperature is increased to\(175^{\circ} \mathrm{C}\) while the pressure is held constant.

A deep-sea diver uses a gas cylinder with a volume of 10.0 \(\mathrm{L}\) and a content of 51.2 \(\mathrm{g}\) of \(\mathrm{O}_{2}\) and 32.6 \(\mathrm{g}\) of He. Calculate the partial pressure of each gas and the total pressure if the temperature of the gas is \(19^{\circ} \mathrm{C}\) .

See all solutions

Recommended explanations on Chemistry 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.

Sign-up for free