Question

A chemist possesses \(1 \mathrm{~cm}^{3}\) of \(\mathrm{O}_{2}\) gas and \(1 \mathrm{~cm}^{3}\) of \(\mathrm{N}_{2}\) gas both at Standard Temperature and Pressure (STP). Compare these gases with respect to (a) number of molecules, and (b) average speed of molecules.

Short answer

(a) The number of \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\) molecules is the same, as they are approximately equal to 2.697 * 10¹⁹ molecules. (b) The average speed of \(\mathrm{N}_{2}\) molecules is greater than the average speed of \(\mathrm{O}_{2}\) molecules, as \(\_{\mathrm{N}_{2}} \approx 514.80\frac{\text{m}}{\text{s}}\) and \(\_{\mathrm{O}_{2}} \approx 480.58\frac{\text{m}}{\text{s}}\).

Step-by-step solution

Calculate the number of moles of each gas

We can use the Ideal Gas Law to find the number of moles of each gas: \(PV=nRT\) At STP, P = 1 atm, V = 0.001 L, and T = 273.15 K. For \(\mathrm{O}_{2}\) gas, we get: \(1 \text{ atm} * 0.001 \text{ L} = n * 0.0821 \frac{\text{L} \cdot \text{atm}}{\text{mol} \cdot \text{K}} * 273.15\text{ K}\) Solve for \(n\,(\mathrm{O}_{2})\): \(n\,(\mathrm{O}_{2}) \approx 4.477 * 10^{-5}\frac{1}{\text{mol}}\) For \(\mathrm{N}_{2}\) gas, we again use the Ideal Gas Law: \(1 \text{ atm} * 0.001 \text{ L} = n * 0.0821 \frac{\text{L} \cdot \text{atm}}{\text{K} \cdot \text{mol}} \cdot 273.15\text{ K}\) Solve for \(n\,(\mathrm{N}_{2})\): \(n\,(\mathrm{N}_{2}) \approx 4.477 *10^{-5}\frac{1}{\text{mol}}\)

Calculate the number of molecules of each gas

Now, we can calculate the number of molecules of each gas using Avogadro's Number (\(6.022*10^{23}\) molecules/mol): For \(\mathrm{O}_{2}\) gas: \(N\,(\mathrm{O}_{2}) = n\,(\mathrm{O}_{2}) * 6.022 * 10^{23} \frac{\text{molecules}}{\text{mol}} \approx 2.697 * 10^{19}\text{ molecules}\) For \(\mathrm{N}_{2}\) gas: \(N\,(\mathrm{N}_{2}) = n\,(\mathrm{N}_{2}) * 6.022 * 10^{23} \frac{\text{molecules}}{\text{mol}} \approx 2.697 * 10^{19}\text{ molecules}\) By comparing the number of molecules for each gas, we find that: (a) The number of \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\) molecules is the same, as they are approximately equal to 2.697 * 10¹⁹ molecules.

Calculate the average speed of molecules for each gas

Now we will calculate the average speed of \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\) molecules using the formula: \(\=\sqrt{\frac{8RT}{\pi M}}\) where R is the gas constant (8.314 J/(mol·K)), T is the temperature, and M is the molar mass of the moving gas molecules. For \(\mathrm{O}_{2}\) gas: \(\_{\mathrm{O}_{2}}=\sqrt{\frac{(8)(8.314 \frac{\text{J}}{\text{mol}\cdot\text{K}})(273.15\text{ K})}{(\pi)(32\frac{\text{kg}}{\text{kmol}})}}\approx 480.58\frac{\text{m}}{\text{s}}\) For \(\mathrm{N}_{2}\) gas: \(\_{\mathrm{N}_{2}}=\sqrt{\frac{(8)(8.314 \frac{\text{J}}{\text{mol}\cdot\text{K}})(273.15\text{ K})}{(\pi)(28\frac{\text{kg}}{\text{kmol}})}}\approx 514.80\frac{\text{m}}{\text{s}}\) By comparing the average speed of \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\) molecules, we find that: (b) The average speed of \(\mathrm{N}_{2}\) molecules is greater than the average speed of \(\mathrm{O}_{2}\) molecules, as \(\_{\mathrm{N}_{2}} \approx 514.80\frac{\text{m}}{\text{s}}\) and \(\_{\mathrm{O}_{2}} \approx 480.58\frac{\text{m}}{\text{s}}\).

Key concepts

Avogadro's Number

Avogadro's Number is a fundamental constant in chemistry, representing the number of constituent particles, usually atoms or molecules, that are contained in one mole of a substance. This number is approximately equal to \(6.022 \times 10^{23}\) particles per mole. When dealing with gases, Avogadro's Number becomes incredibly useful as it allows chemists to convert between the number of moles and the number of molecules. For example, if you know the amount of gas in moles, you can multiply by Avogadro's Number to find out how many molecules you have. This conversion is essential in comparing quantities of gases at the molecular level as shown in the solution where the number of molecules of \(\mathrm{O}_2\) and \(\mathrm{N}_2\) is determined using this constant.

Standard Temperature and Pressure

Standard Temperature and Pressure (STP) is a reference point used in chemistry to provide a basis for comparisons of gases. STP is defined as a temperature of \(273.15\) Kelvin (\(0\) degrees Celsius) and a pressure of \(1\) atmosphere (atm). These conditions provide a standard baseline that makes it easier to use the Ideal Gas Law, which describes the behavior of gases under various conditions. At STP, the behavior of gases is predictable, allowing for straightforward calculations of their properties, such as volume, pressure, and temperature relationships. The problem you've seen utilizes STP to ensure consistent calculations for gas volumes and to find the number of moles present using the Ideal Gas Law.

Molecular Speed

Molecular speed refers to the average speed at which molecules of a gas travel at a given temperature. The kinetic theory of gases states that the temperature of a gas is directly related to the average kinetic energy of its molecules. As the temperature increases, the molecular speed increases as well. The formula for calculating the average speed of molecules, \(\\), at a specific temperature is \(\ = \sqrt{\frac{8RT}{\pi M}}\), where \(R\) is the gas constant, \(T\) is the temperature in Kelvin, and \(M\) is the molar mass of the gas. The exercise uses this formula to compare the average speeds of \(\mathrm{O}_2\) and \(\mathrm{N}_2\) at STP. The difference in their speeds can be attributed to their differing molar masses, with \(\mathrm{N}_2\) molecules moving faster than \(\mathrm{O}_2\) molecules.

Molar Mass

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole \((\text{g/mol})\). It is a crucial property for determining how much of a substance is involved in a chemical reaction or calculation. For gases, molar mass directly affects molecular speed and other properties like density. When considering the Ideal Gas Law, molar mass helps in understanding how gases will behave under various conditions. In the exercise, molar mass is essential for calculating the average molecular speed, with lighter \(\mathrm{N}_2\) molecules (molar mass of \(28 \text{kg/kmol}\)) moving faster than heavier \(\mathrm{O}_2\) molecules (molar mass of \(32 \text{kg/kmol}\)). Understanding molar mass is fundamental to interpreting and predicting the behavior of different gases in various contexts.