Question

Consider separate \(1.0\) -L gaseous samples of \(\mathrm{He}, \mathrm{N}_{2}\), and \(\mathrm{O}_{2}\), all at STP and all acting ideally. Rank the gases in order of increasing average kinetic energy and in order of increasing average velocity.

Short answer

The average kinetic energy of the given gases (He, N₂, O₂) is the same because they are all at STP and acting ideally. Therefore, the ranking based on average kinetic energy is He = N₂ = O₂. However, their average velocities are different due to their varying molar masses. The ranking based on average velocity is O₂ < N₂ < He.

Step-by-step solution

Calculate the average kinetic energy for each gas

Since the gases are at STP and are acting ideally, their average kinetic energy is directly proportional to their temperature. Recall that: Average Kinetic Energy \(= \frac{3}{2} k T\) where \(k\) is Boltzmann's constant (\(1.38 \times 10^{-23} J K^{-1}\)) and \(T\) is the temperature (273.15 K). Since the temperature is the same for all three gases, their average kinetic energies are also the same. Ranking based on average kinetic energy: He = N₂ = O₂

Calculate the average velocity for each gas

Now we need to find the average velocities of these gases and rank them. Recall that the formula for average velocity (\(v\)) is as follows: \(v = \sqrt{\frac{2 K_{avg}}{m}}\) where \(K_{avg}\) is the gas's average kinetic energy and \(m\) is the molar mass of the gas. From Step 1, we know that the average kinetic energies of all gases are the same. However, molar masses are different for these gases: \(4 g/mol\) for He, \(28 g/mol\) for N₂, and \(32 g/mol\) for O₂. Now, since the average kinetic energy is the same for all gases, we can say that: \(\frac{v_1}{v_2} = \sqrt{\frac{m_2}{m_1}}\) Comparing He (1) and N₂ (2), \(\frac{v_{He}}{v_{N₂}} = \sqrt{\frac{28}{4}}\) Comparing He (1) and O₂ (3), \(\frac{v_{He}}{v_{O₂}} = \sqrt{\frac{32}{4}}\) Comparing N₂ (2) and O₂ (3), \(\frac{v_{N₂}}{v_{O₂}} = \sqrt{\frac{32}{28}}\) Solving these equations, we get: \(v_{He} > v_{N₂} > v_{O₂}\) Ranking based on average velocity: O₂ < N₂ < He

Key concepts

STP Conditions

Understanding Standard Temperature and Pressure (STP) conditions is crucial in the study of gas behavior. These conditions serve as a reference point for many experiments and calculations in chemistry and physics. STP refers to a temperature of 0°C or 273.15 Kelvin and a pressure of 1 atmosphere (atm). At STP, one mole of an ideal gas occupies 22.4 liters of volume. This is key because under these conditions, the behavior of gases can be predicted with greater simplicity and accuracy based on the ideal gas law.

For students, grasping the concept of STP is essential when comparing properties of different gases, as it ensures that a standard baseline is used. It's like comparing the performance of different athletes when they are all subject to the same conditions. By using STP, we can directly compare the properties of different gases, such as helium (He), nitrogen (N₂), and oxygen (O₂), without the complication of varying temperatures and pressures affecting the results.

Average Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion, and when talking about gases, we refer to the average kinetic energy of the gas particles. The kinetic molecular theory explains that the average kinetic energy of a gas is directly related to its temperature. This relationship is beautifully expressed by the equation:

\[ \text{Average Kinetic Energy} = \frac{3}{2} k T \]

Here, \( k \) is the Boltzmann's constant, and \( T \) is the absolute temperature in Kelvin. At STP conditions, since the temperature is constant, the average kinetic energy of any ideal gas is the same. This means that for gaseous samples of He, N₂, and O₂ at STP, they all share the same average kinetic energy. Therefore, they are all 'equally energetic' on average, regardless of the gas type. For students, it's like thinking of different cars with engines running at the same RPM on a neutral gear; they're all putting out similar power, though they might differ in other properties such as mass.

Root Mean Square Velocity

The root mean square velocity (RMS velocity) is a way to express the average velocity of gas particles. When tackling gaseous behavior, it’s not just the average kinetic energy we are interested in but also how fast the particles are moving on average. RMS velocity is important because it relates the kinetic energy of gas particles to their masses and temperatures.

The formula for RMS velocity is given by:

\[ v = \sqrt{\frac{2 K_{\text{avg}}}{m}} \]

Where \( v \) is the RMS velocity, \( K_{\text{avg}} \) is the average kinetic energy (which is same for all gases at STP as established earlier), and \( m \) is the molar mass of the gas. From this relationship, we understand that at a constant kinetic energy, lighter gas molecules will move faster. This is because RMS velocity is inversely proportional to the square root of the molar mass. Consequently, helium atoms, being the lightest, move the fastest, followed by nitrogen and oxygen molecules. To visualize this, imagine a group of runners: a lightweight athlete can sprint faster than heavier ones if all have the same amount of kinetic energy — it’s about how that energy translates to motion.