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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

Calculate the change in internal energy of 1.00 mole of a diatomic ideal gas that starts at room temperature \((293 \mathrm{~K})\) when its temperature is increased by \(2.00 \mathrm{~K}\).

Two identical containers hold equal masses of gas, oxygen in one and nitrogen in the other. The gases are held at the same temperature. How does the pressure of the oxygen compare to that of the nitrogen? a) \(p_{\mathrm{O}}>p_{\mathrm{N}}\) b) \(p_{\mathrm{O}}=p_{\mathrm{N}}\) c) \(p_{\mathrm{O}}

Chapter 13 examined the variation of pressure with altitude in the Earth's atmosphere, assuming constant temperature-a model known as the isothermal atmosphere. A better approximation is to treat the pressure variations with altitude as adiabatic. Assume that air can be treated as a diatomic ideal gas with effective molar mass \(M_{\text {air }}=28.97 \mathrm{~g} / \mathrm{mol}\) a) Find the air pressure and temperature of the atmosphere as functions of altitude. Let the pressure at sea level be \(p_{0}=101.0 \mathrm{kPa}\) and the temperature at sea level be \(20.0^{\circ} \mathrm{C}\) b) Determine the altitude at which the air pressure and density are half their sea-level values. What is the temperature at this altitude, in this model? c) Compare these results with the isothermal model of Chapter \(13 .\)

When you blow hard on your hand, it feels cool, but when you breathe softly, it feels warm. Why?

A monatomic ideal gas expands isothermally from \(\left\\{p_{1}, V_{1}, T_{1}\right\\}\) to \(\left\\{p_{2}, V_{2}, T_{1}\right\\} .\) Then it undergoes an isochoric process, which takes it from \(\left\\{p_{2}, V_{2}, T_{1}\right\\}\) to \(\left\\{p_{1}, V_{2}, T_{2}\right\\}\) Finally the gas undergoes an isobaric compression, which takes it back to \(\left\\{p_{1}, V_{1}, T_{1}\right\\}\) a) Use the First Law of Thermodynamics to find \(Q\) for each of these processes. b) Write an expression for total \(Q\) in terms of \(p_{1}, p_{2}, V_{1},\) and \(V_{2}\).
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