Problem 69 - University Physics with Modern Physics

Question

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

Step-by-Step Solution

Answer

Answer: The total mass of oxygen molecules in a cubic meter of air at normal temperature and pressure is approximately 273.92 grams.

Step 1: 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.

Step 2: 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.

Step 3: 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: \(n = \dfrac{(1.01 \times 10^5 Pa) (0.21m^3)}{(8.314 J/(mol \cdot K))(298.15 K)}\ \approx 8.56\ \text{moles}\)

Step 4: 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.

Key Concepts

Oxygen Molecules

At room temperature and pressure, we often consider the composition of air to understand various properties and reactions. Oxygen molecules, represented as \(\mathrm{O}_2\), are a significant component of our atmosphere. They make up about 21% by volume of the air we breathe. This diatomic molecule, \(\mathrm{O}_2\), consists of two oxygen atoms bonded together. This configuration is important in many biological and chemical processes, such as respiration and combustion. A key point is that in various calculations, such as those involving gas laws, we treat the ozone as individual units of \(\mathrm{O}_2\) without breaking them down into single oxygen atoms.

Normal Temperature and Pressure

In scientific contexts, when we refer to normal temperature and pressure (NTP), we typically mean a standard state where the temperature is 25°C (or 298.15 Kelvin), and the pressure is 1.01 x \(10^5\) PascaIs. This is a common condition used to simplify calculations and is also referred to as "room temperature" in everyday parlance.Understanding NTP is crucial when calculating the behaviors of gases, as it provides a consistent frame of reference. Under these conditions, the density and volume of gases can be accurately characterized, enabling precise experimentation and application in various scientific and industrial processes.

Moles Calculation

Calculating the number of moles of a substance is an essential step in understanding chemical quantities. In the Ideal Gas Law, represented as \(PV = nRT\), the quantity \(n\) refers to moles of gas.First, ensure you have the volume (\(V\)), pressure (\(P\)), and temperature (\(T\)) in the correct units: meters cubed for volume, Pascals for pressure, and Kelvin for temperature. Using our problem parameters:- Volume of oxygen (\(V\)) = 0.21 \(m^3\) (21% of the volume is oxygen)- Pressure (\(P\)) = 1.01 x \(10^5\) Pa- Temperature (\(T\)) = 298.15 KWe rearrange the Ideal Gas Law to solve for moles \(n = \frac{P \cdot V}{R \cdot T}\), where \(R\) is 8.314 J/(mol·K). This application will reveal the number of moles in the given volume of oxygen under these conditions.

Molar Mass

Molar mass is a central concept in chemistry and represents the mass of one mole of a given substance. For oxygen molecules (\(\mathrm{O}_2\)), the molar mass is 32 g/mol. This is because each oxygen atom has a molar mass of approximately 16 g/mol, and a molecule contains two of these atoms. The molar mass allows us to convert between the grams of a substance and its moles, which is helpful for quantitative analysis in chemistry. For instance, knowing the moles of oxygen we calculated using the Ideal Gas Law, we convert these moles into a mass by multiplying the number of moles by the molar mass. This step allows us to determine the total mass of oxygen in grams within the specified volume and conditions. Thus, understanding molar mass simplifies many complex chemical processes by bridging the gap between the laboratory scale and atomic scale measurements.