Question

Suppose you have two 1 -L flasks, one containing \(\mathrm{N}_{2}\) at STP, the other containing \(\mathrm{CH}_{4}\) at STP. How do these systems compare with respect to (a) number of molecules, (b) density, (c) average kinetic energy of the molecules, (d) rate of effusion through a pinhole leak?

Short answer

a) Both N2 and CH4 have the same number of molecules in a 1-L flask at STP. b) N2 is denser than CH4 (\(1.25\,\text{g/L}\) vs \(0.713\,\text{g/L}\)). c) The average kinetic energy of the molecules is the same for both N2 and CH4 at STP. d) The rate of effusion of CH4 is greater than that of N2.

Step-by-step solution

a) Number of molecules

First, we need to find the number of moles of each gas in a 1-L flask at STP. Since 1 mole of an ideal gas occupies 22.4 L at STP, the number of moles in a 1-L flask is given by: \(n = \frac{1 \text{ L}}{22.4 \text{ L/mol}} = 0.04464 \text{ mol}\) Using Avogadro's number, we convert the moles to the number of molecules: \(\text{molecules} = n \times Avogadro's\, number\) \(\text{molecules} = 0.04464 \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol}\) The number of molecules in each flask is the same, because the number of moles is the same for both N2 and CH4 at STP.

b) Density

To find the density, we need to determine the mass of each gas with the given molecular weights. Density = mass / volume For N2: \(\rho_\text{N2} = \frac{0.04464\,\text{mol} \times 28\,\text{g/mol}}{1\,\text{L}} = 1.25\,\text{g/L}\) For CH4: \(\rho_\text{CH4} = \frac{0.04464\,\text{mol} \times 16\,\text{g/mol}}{1\,\text{L}} = 0.713\,\text{g/L}\) So, N2 is denser than CH4.

c) Average kinetic energy

The formula for average kinetic energy (Ek) of a molecule in a gas: \(Ek = \frac{3}{2}kT\) Where k is the Boltzmann constant, k = \(1.38 \times 10^{-23} J/K\), and T is the temperature in Kelvin. For both N2 and CH4, the Ek is the same at STP: \(Ek = \frac{3}{2}(1.38 \times 10^{-23}\,\text{J/K}) \times 273.15\,\text{K}\) The average kinetic energy of the molecules is the same for both N2 and CH4 at STP.

d) Rate of effusion

To find the rate of effusion, we use Graham's law: \(Rate_1 / Rate_2 = \sqrt{M_2 / M_1}\) Comparing N2 (Rate_1) and CH4 (Rate_2): \(\frac{\text{Rate}_\text{N2}}{\text{Rate}_\text{CH4}} = \sqrt{\frac{16\,\text{g/mol}}{28\,\text{g/mol}}}\) Thus, the rate of effusion of CH4 is greater than that of N2. In summary: a) The number of molecules is the same for both N2 and CH4 at STP. b) N2 is denser than CH4. c) The average kinetic energy of the molecules is the same for both N2 and CH4 at STP. d) The rate of effusion of CH4 is greater than that of N2.

Key concepts

Molecular Density

Molecular density is the mass of molecules within a given volume. It is vital to understand this when comparing gases at standard temperature and pressure (STP). At STP, 1 mole of any gas occupies 22.4 liters. Thus, in a 1-liter container, the number of moles will be approximately 0.04464 moles. The molecular density helps us comprehend how tightly packed the molecules are within a given space.
To calculate molecular density, the formula is used: \( \text{Density} = \frac{\text{mass}}{\text{volume}} \). For each gas, you must know the molecular weight to determine its mass. For nitrogen (N\(_2\)), the molecular weight is 28 g/mol, whereas for methane (CH\(_4\)), it is 16 g/mol.

• For N\(_2\), density is \( \frac{28 \times 0.04464}{1} = 1.25 \, \text{g/L} \).
• For CH\(_4\), density is \( \frac{16 \times 0.04464}{1} = 0.713 \, \text{g/L} \).

This indicates that at STP, N\(_2\) is denser than CH\(_4\). Understanding molecular density is key in various applications, including chemical reactions and gas behavior under different conditions.

Kinetic Theory of Gases

The kinetic theory of gases is fundamental in understanding the behavior of gases at molecular levels. It provides a model that helps visualize and calculate the effects of variables such as pressure, volume, and temperature. This theory suggests that gas molecules are in constant, random motion and that their interactions are perfectly elastic collisions, meaning no energy is lost.
The average kinetic energy of gas molecules is directly proportional to the temperature, as expressed by the equation:\[ E_k = \frac{3}{2} kT \]where \( E_k \) is the average kinetic energy, \( k \) is the Boltzmann constant \( (1.38 \times 10^{-23} \, \text{J/K}) \), and \( T \) is the temperature in Kelvin.
At STP, the temperature is 273.15 K, leading to the same average kinetic energy for different gases. This means that regardless of the gas type or mass, they have equivalent average kinetic energy at the same temperature. This concept helps explain gas expansion, compression, and molecular movement efficiently by using a temperature scale to determine motion intensity.

Graham's Law of Effusion

Graham's law of effusion describes how gases escape through a tiny hole into a vacuum without external pressure differences. It indicates that the rate of effusion is inversely proportional to the square root of the gas's molar mass.
The formula used is:\[ \frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{M_2}{M_1}} \]Here, \( \text{Rate}_1 \) and \( \text{Rate}_2 \) are the effusion rates of two different gases, while \( M_1 \) and \( M_2 \) are their molar masses.
In the case of nitrogen (N\(_2\)) and methane (CH\(_4\)), nitrogen has a molar mass of 28 g/mol and methane 16 g/mol. Therefore, using Graham's law:\[ \frac{\text{Rate}_{\text{N}_2}}{\text{Rate}_{\text{CH}_4}} = \sqrt{\frac{16}{28}} \]This indicates that CH\(_4\) effuses more rapidly than N\(_2\). As a result, gases with lighter molecules move faster and have higher effusion rates. This concept is crucial when considering gas diffusion in rooms or measuring gas flow through porous materials.