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Chapter 19: Problem 68

What is the total mass of all the oxygen molecules in a cubic meter of air at normal temperature \(\left(25^{\circ} \mathrm{C}\right)\) and pressure \(\left(1.01 \cdot 10^{5} \mathrm{~Pa}\right) ?\) Note that air is about \(21 \%\) (by volume) oxygen (molecular \(\mathrm{O}_{2}\) ), with the remainder being primarily nitrogen (molecular \(\mathrm{N}_{2}\) ).

Short Answer

Expert verified
Answer: The total mass of oxygen molecules in a cubic meter of air at normal temperature and pressure is approximately 273.6 grams.

Step by step solution

01

Calculate the number of moles of air in a cubic meter

To calculate the number of moles in a cubic meter of air, we can use the ideal gas equation: \(PV = nRT\) where P is the pressure (in Pa), V is the volume (in m³), n is the number of moles, R is the ideal gas constant (8.314 J/mol·K), and T is the temperature (in Kelvin) First, we need to convert the temperature from Celsius to Kelvin: \(T = 25^{\circ}C + 273.15 = 298.15 K\) Now, we can plug in the known values and solve for n: \((1.01 \cdot 10^5 Pa) \cdot (1 m^3) = n \cdot (8.314 J/mol·K) \cdot (298.15 K)\) Solving for n, we get: \(n = \dfrac{1.01 \cdot 10^5 Pa \cdot 1 m^3}{8.314 J/mol·K \cdot 298.15 K} \approx 40.73 \, mol\)
02

Find the proportion of oxygen in the air

We are given that air is 21% oxygen by volume. Therefore, to find the number of moles of oxygen in 1 cubic meter of air, we can multiply the total moles (40.73) by 0.21 (21%): \(n_{O_2} = 40.73 \, mol \cdot 0.21 = 8.55 \, mol\)
03

Calculate the mass of oxygen in a cubic meter of air

Now that we have the number of moles of oxygen, we can calculate the mass using the molecular weight of oxygen (O₂): Molecular weight of O₂ = 2 * molecular weight of O = 2 * 16 g/mol = 32 g/mol. The mass of oxygen (m_O₂) can be calculated using the number of moles and the molecular weight: \(m_{O_2} = n_{O_2} \cdot MW_{O_2} = 8.55 \, mol \cdot 32 \dfrac{g}{mol} = 273.6 g\) So, the total mass of oxygen molecules in a cubic meter of air at normal temperature and pressure is approximately 273.6 grams.

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Most popular questions from this chapter

Liquid bromine \(\left(\mathrm{Br}_{2}\right)\) is spilled in a laboratory accident and evaporates. Assuming the vapor behaves as an ideal gas, with a temperature \(20.0^{\circ} \mathrm{C}\) and a pressure of \(101.0 \mathrm{kPa}\), find its density.

A sample of gas at \(p=1000 . \mathrm{Pa}, V=1.00 \mathrm{~L},\) and \(T=300 . \mathrm{K}\) is confined in a cylinder. a) Find the new pressure if the volume is reduced to half of the original volume at the same temperature. b) If the temperature is raised to \(400 . \mathrm{K}\) in the process of part (a), what is the new pressure? c) If the gas is then heated to \(600 . \mathrm{K}\) from the initial value and the pressure of the gas becomes \(3000 . \mathrm{Pa},\) what is the new volume?

Which of the following gases has the highest rootmean-square speed? a) nitrogen at \(1 \mathrm{~atm}\) and \(30^{\circ} \mathrm{C}\) b) argon at \(1 \mathrm{~atm}\) and \(30^{\circ} \mathrm{C}\) c) argon at \(2 \mathrm{~atm}\) and \(30^{\circ} \mathrm{C}\) d) oxygen at 2 atm and \(30^{\circ} \mathrm{C}\) e) nitrogen at \(2 \mathrm{~atm}\) and \(15^{\circ} \mathrm{C}\)

An ideal gas has a density of \(0.0899 \mathrm{~g} / \mathrm{L}\) at \(20.00^{\circ} \mathrm{C}\) and \(101.325 \mathrm{kPa}\). Identify the gas.

6.00 liters of a monatomic ideal gas, originally at \(400 . \mathrm{K}\) and a pressure of \(3.00 \mathrm{~atm}\) (called state 1 ), undergo the following processes: \(1 \rightarrow 2\) isothermal expansion to \(V_{2}=4 V_{1}\) \(2 \rightarrow 3\) isobaric compression \(3 \rightarrow 1\) adiabatic compression to its original state Find the pressure, volume, and temperature of the gas in states 2 and \(3 .\) How many moles of the gas are there?
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