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Chapter 19: Problem 69

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 273.92 grams.

Step by step solution

01

Write the Ideal Gas Law equation

The Ideal Gas Law equation is given by: \(PV=nRT\) Where: P is the pressure of the gas, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. We will use this equation to find the number of moles of oxygen in the air.
02

Convert temperature to Kelvin

In order to work with the Ideal Gas Law equation, we need to have the temperature in Kelvin. To do this, we add 273.15 to the Celsius temperature: \(T_{K} = T_{C} + 273.15\) \(T = 25^{\circ}C + 273.15 = 298.15K\) Now, we have the temperature in Kelvin.
03

Calculate the number of moles of oxygen

As given, air is about \(21\%\) oxygen. Therefore, the volume of oxygen in a cubic meter of air is: \(V_{O_2} = 0.21 \times 1m^3 = 0.21m^3\) Now, we can use the Ideal Gas Law equation to calculate the number of moles of oxygen: \((1.01 \times 10^5 Pa)(0.21m^3)=(n)(8.314 J/(mol \cdot K))(298.15 K)\) Solving for n, we get: \(n = \dfrac{(1.01 \times 10^5 Pa) (0.21m^3)}{(8.314 J/(mol \cdot K))(298.15 K)}\ \approx 8.56\ \text{moles}\)
04

Convert moles to mass

To find the total mass of oxygen, we multiply the number of moles by the molar mass of oxygen gas: \(Mass_{O_2} = (8.56\ \text{moles})(32 g/mol) \approx 273.92\ \text{grams}\) Therefore, the total mass of oxygen molecules in a cubic meter of air at normal temperature and pressure is approximately 273.92 grams.

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Most popular questions from this chapter

At Party City, you purchase a helium-filled balloon with a diameter of \(40.0 \mathrm{~cm}\) at \(20.0^{\circ} \mathrm{C}\) and at \(1.00 \mathrm{~atm} .\) a) How many helium atoms are inside the balloon? b) What is the average kinetic energy of the atoms? c) What is the root-mean-square speed of the atoms?

Treating air as an ideal gas of diatomic molecules, calculate how much heat is required to raise the temperature of the air in an \(8.00 \mathrm{~m}\) by \(10.0 \mathrm{~m}\) by \(3.00 \mathrm{~m}\) room from \(20.0^{\circ} \mathrm{C}\) to \(22.0^{\circ} \mathrm{C}\) at \(101 \mathrm{kPa}\). Neglect the change in the number of moles of air in the room.

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?

In a period of \(6.00 \mathrm{~s}, 9.00 \cdot 10^{23}\) nitrogen molecules strike a section of a wall with an area of \(2.00 \mathrm{~cm}^{2}\). If the molecules move with a speed of \(400.0 \mathrm{~m} / \mathrm{s}\) and strike the wall head on in elastic collisions, what is the pressure exerted on the wall? (The mass of one \(\mathrm{N}_{2}\) molecule is \(4.68 \cdot 10^{-26} \mathrm{~kg}\).)

At a temperature of \(295 . \mathrm{K},\) the vapor pressure of pentane \(\left(\mathrm{C}_{5} \mathrm{H}_{12}\right)\) is \(60.7 \mathrm{kPa}\). Suppose \(1.000 \mathrm{~g}\) of gaseous pentane is contained in a cylinder with diathermal (thermally conducting) walls and a piston to vary the volume. The initial volume is \(1.000 \mathrm{~L},\) and the piston is moved in slowly, keeping the temperature at \(295 \mathrm{~K}\). At what volume will the first drop of liquid pentane appear?
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