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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 m鲁), n is the number of moles, R is the ideal gas constant (8.314 J/mol路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: \((1.01 \cdot 10^5 Pa) \cdot (1 m^3) = n \cdot (8.314 J/mol路K) \cdot (298.15 K)\) Solving for n, we get: \(n = \dfrac{1.01 \cdot 10^5 Pa \cdot 1 m^3}{8.314 J/mol路K \cdot 298.15 K} \approx 40.73 \, mol\)
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 (O鈧�): Molecular weight of O鈧� = 2 * molecular weight of O = 2 * 16 g/mol = 32 g/mol. The mass of oxygen (m_O鈧�) 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

The kinetic theory of an ideal gas takes into account not only translational motion of atoms or molecules but also, for diatomic and polyatomic gases, vibration and rotation. Will the temperature increase from a given amount of energy being supplied to a monatomic gas differ from the temperature increase due to the same amount of energy being supplied to a diatomic gas? Explain.

Air at 1.00 atm is inside a cylinder \(20.0 \mathrm{~cm}\) in radius and \(20.0 \mathrm{~cm}\) in length that sits on a table. The top of the cylinder is sealed with a movable piston. A \(20.0-\mathrm{kg}\) block is dropped onto the piston. From what height above the piston must the block be dropped to compress the piston by \(1.00 \mathrm{~mm} ? 2.00 \mathrm{~mm} ? 1.00 \mathrm{~cm} ?\)

Calculate the root-mean-square speed of air molecules at room temperature \(\left(22.0^{\circ} \mathrm{C}\right)\) from the kinetic theory of an ideal gas.

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 \(\mathrm{N}_{2}\) ).

The electrons in a metal that produce electric currents behave approximately as molecules of an ideal gas. The mass of an electron is \(m_{\mathrm{e}} \doteq 9.109 \cdot 10^{-31} \mathrm{~kg} .\) If the temperature of the metal is \(300.0 \mathrm{~K},\) what is the root-mean-square speed of the electrons?
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