Question

Suppose you have two 1 -L flasks, one containing \(N_{2}\) 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 flasks have the same number of molecules since they have the same volume, temperature, and pressure conditions. (b) The density of N2 is greater than the density of CH4 due to their respective molar masses. (c) The average kinetic energy of the molecules in both flasks is the same since they have the same temperature. (d) The rate of effusion of CH4 is greater than the rate of effusion of N2 due to their respective molar masses and Graham's law of effusion.

Step-by-step solution

Number of molecules

To find the number of molecules in each flask, we can use the ideal gas law, which is given as: \[PV=nRT\] where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. Since both gases are at STP, we have: P = 1 atm (standard pressure) V = 1 L (volume of each flask) T = 273.15 K (standard temperature) R = 0.0821 L atm/(mol K) (ideal gas constant) Now we will find the number of moles (n) in each flask: \[n = \frac{PV}{RT}\] For both N2 and CH4, we have the same temperature, pressure, and volume, so the number of moles will be equal. Thus, by extension, the number of molecules in each flask will also be equal since the number of molecules is equal to the number of moles multiplied by Avogadro's number (6.022 x 10^23).

Density

To find the density of each flask, we can use the formula: \[Density = \frac{Mass}{Volume}\] We can find the mass of each gas using the number of moles (n) multiplied by the molar mass (M) of each gas: \[Mass = nM\] Molar mass of N2 = 28 g/mol Molar mass of CH4 = 16 g/mol Now we can find the density for each gas: Plotting the values we have calculated before, we find out that the density of N2 is greater than the density of CH4.

Average kinetic energy of the molecules

The average kinetic energy of the molecules is given by the formula: \[KE_{avg} = \frac{3}{2}kT\] where k is the Boltzmann constant (1.38 x 10^(-23) J/K) and T is the temperature. Since both N2 and CH4 are at STP, they have the same temperature, so their average kinetic energy will be the same.

Rate of effusion through a pinhole leak

To find the rate of effusion, we can use Graham's law of effusion, which is given by: \[\frac{Rate_{1}}{Rate_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}\] Where Rate1 and Rate2 are the effusion rates of the two gases, and M1 and M2 are their molar masses. We are given that one flask contains N2 and the other contains CH4. Using the molar masses we found in step 2, we have: M1 = 28 g/mol (N2) M2 = 16 g/mol (CH4) Now, we can find the ratio of their effusion rates: \[\frac{Rate_{N_{2}}}{Rate_{CH_{4}}} = \sqrt{\frac{16}{28}}\] We find that the rate of effusion of CH4 is greater than the rate of effusion of N2. In summary: - The number of molecules is equal in both flasks. - The density of N2 is greater than the density of CH4. - The average kinetic energy of the molecules in both flasks is the same. - The rate of effusion of CH4 is greater than the rate of effusion of N2.

Key concepts

STP (Standard Temperature and Pressure)

When discussing gas behavior, it's crucial to consider the conditions under which the gas is measured. STP, which stands for Standard Temperature and Pressure, refers to a set of conditions where a gas is at 0 degrees Celsius (273.15 Kelvin) and 1 atmosphere of pressure. This standard reference point is essential in chemistry to compare the properties of gases, such as volume, density, and rate of effusion.

At STP, one mole of an ideal gas occupies 22.4 liters, a volume known as the molar volume. In the exercise provided, STP allows us to assume the gases in question have the same volume and temperature, simplifying calculations such as understanding the number of molecules and their behavior. Remembering that STP is a simplified model, in reality, gases might deviate from this behavior, particularly at high pressures or low temperatures where ideal gas laws may no longer strictly apply.

Molecular Kinetic Energy

Molecular kinetic energy represents the energy that gas molecules possess due to their motion. The average kinetic energy of a gas molecule is directly proportional to the absolute temperature of the gas. In equation form, the average kinetic energy is given by \(KE_{avg} = \frac{3}{2}kT\), where \(k\) is the Boltzmann constant and \(T\) is the temperature in Kelvin.

In the scenario given, both the \(N_{2}\) and \(CH_{4}\) gases are at STP, which means they are at the same temperature. Therefore, they share the same average kinetic energy, regardless of the differences in their molecular weights. This is an important concept in thermodynamics, reflecting that temperature is a measure of average kinetic energy of the molecules within a substance.

Rate of Effusion

Effusion is the process by which gas molecules escape through a tiny opening into a vacuumed environment. The rate of effusion is influenced by the gas's temperature and the mass of its molecules. Lighter gas molecules will effuse more rapidly than heavier ones when all other conditions are identical.

To compare effusion rates between two gases, like \(N_{2}\) and \(CH_{4}\) in our exercise, we look at their respective molar masses. The lower the molar mass, the faster the gas effuses. This is due to the molecular kinetic energy being similar for both gases at a given temperature; lighter molecules move faster than heavier ones. The exercise demonstrates how to calculate the relative rates of effusion and concludes that \(CH_{4}\), being lighter, effuses more quickly than \(N_{2}\).

Graham's Law

Graham's law quantitatively describes the relationship between the rate of effusion and the molar masses of gases. It states that the rate of effusion for a gas is inversely proportional to the square root of its molar mass. Mathematically, it is expressed as \(\frac{Rate_{1}}{Rate_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}\), where \(Rate_{1}\) and \(Rate_{2}\) are the effusion rates and \(M_{1}\) and \(M_{2}\) are the molar masses of gas 1 and gas 2, respectively.

In practice, as seen in the exercise provided, Graham's law allows us to predict that methane (\(CH_{4}\)) will effuse faster than nitrogen (\(N_{2}\)) due to its lower molar mass. Understanding Graham's law is fundamental when studying gas behavior and provides insights into applications such as gas separation techniques, respiratory physiology, and even the design of leak detection systems.