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Chapter 10: Problem 38

(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 \(1.95 \times 10^{16}\) CO鈧� 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 m鲁. 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 脳 6.022 x \(10^{23}\) ozone molecules/mol Number of ozone molecules 鈮� \(1.81 \times 10^{18}\) ozone molecules. Answer for Part (a): There are approximately \(1.81 \times 10^{18}\) ozone molecules in a liter. (b) Number of CO鈧� 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 掳C, which is 300 K), and volume (2.0 L, which is 0.002 m鲁) 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 脳 0.002) / (0.0821 脳 300) n 鈮� 8.1 脳 \(10^{-5}\) moles of air
07

Calculate the moles of CO鈧� in the sample

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

Calculate the number of CO鈧� molecules

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

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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鈥檚 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路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.

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Most popular questions from this chapter

Natural gas is very abundant in many Middle Eastern oil fields. However, the costs of shipping the gas to markets in other parts of the world are high because it is necessary to liquefy the gas, which are high because it is necessary to point at atmospheric pressure of \(-164^{\circ} \mathrm{C} .\) One possible strategy is to oxidize the methane to methanol, CH \(_{3} \mathrm{OH}\) , which has a boiling point of \(65^{\circ} \mathrm{C}\) and can therefore be shipped more readily. Suppose that \(10.7 \times 10^{9} \mathrm{ft}^{3}\) of methane at atmospheric pressure and \(25^{\circ} \mathrm{C}\) is oxidized to methanol. (a) What volume of methanol is formed if the density of \(\mathrm{CH}_{3} \mathrm{OH}\) is 0.791 \(\mathrm{g} / \mathrm{mL} ?\) (b) Write balanced chemical equations for the oxidations of methane and methanol to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l) .\) Calculate the total enthalpy change for complete combustion of the \(10.7 \times 10^{9} \mathrm{ft}^{3}\) of methane just described and for complete combustion of the equivalent amount of methanol, as calculated in part (a). (c) Methane, when liquefied, has a density of 0.466 \(\mathrm{g} / \mathrm{mL}\) ; the density of methanol at \(25^{\circ} \mathrm{C}\) is 0.791 \(\mathrm{g} / \mathrm{mL} .\) Compare the enthalpy change upon combustion of a unit volume of liquid methane and liquid methanol. From the standpoint of energy production, which substance has the higher enthalpy of combustion per unit volume?

Mars has an average atmospheric pressure of 0.007 atm. Would it be easier or harder to drink from a straw on Mars than on Earth? Explain. [Section 10.2]

A sample of 3.00 \(\mathrm{g}\) of \(\mathrm{SO}_{2}(g)\) originally in a 5.00 -L vessel at \(21^{\circ} \mathrm{C}\) is transferred to a \(10.0-\mathrm{L}\) vessel at \(26^{\circ} \mathrm{C} .\) A sample of 2.35 \(\mathrm{g}\) of \(\mathrm{N}_{2}(g)\) originally in a \(2.50-\mathrm{L}\) vessel at \(20^{\circ} \mathrm{C}\) is transferred to this same 10.0 -L vessel. (a) What is the partial pressure of \(S O_{2}(g)\) in the larger container? (b) What is the partial pressure of \(N_{2}(g)\) in this vessel? (c) What is the total pressure in the vessel?

A rigid vessel containing a \(3 : 1\) mol ratio of carbon dioxide and water vapor is held at \(200^{\circ} \mathrm{C}\) where it has a total pressure of 2.00 atm. If the vessel is cooled to \(10^{\circ} \mathrm{C}\) so that all of the water vapor condenses, what is the pressure of carbon dioxide? Neglect the volume of the liquid water that forms on cooling.

The molar mass of a volatile substance was determined by the Dumas-bulb method described in Exercise \(10.53 .\) The unknown vapor had a mass of 0.846 g; the volume of the bulb was \(354 \mathrm{cm}^{3},\) pressure 752 torr, and temperature \(100^{\circ} \mathrm{C}\) . Calculate the molar mass of the unknown vapor.
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