Question

Two non-reactive monoatomic ideal gases have their atomic masses in the ratio \(2: 3\). The ratio of their partial pressures, when enclosed in a vessel kept at a constant temperature, is \(4: 3\). The ratio of their densities is (A) \(1: 4\) (B) \(1: 2\) (C) \(6: 9\) (D) \(8: 9\)

Short answer

The ratio of their densities is (D) \(8:9\).

Step-by-step solution

- Recall the Ideal Gas Law

For ideal gases, the Ideal Gas Law applies: \(PV = nRT\). Here, \(P\) 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 the temperature is constant and the gases are in the same vessel, \(V\) and \(T\) are the same for both gases.

- Relate Pressure to Moles

For each gas, the pressure can be related to the number of moles and thus to the mass and molar mass: \(P = \frac{nRT}{V} = \frac{(mass/molar mass)RT}{V}\). Given that the atomic masses are in the ratio \(2:3\), the molar masses are in the same ratio.

- Calculate Density Ratio

The density \(\rho\) of a substance is its mass per unit volume, \(\rho = mass/V\). Using the Ideal Gas Law rearranged for mass, we find that \(mass = \frac{PVmolar mass}{RT}\). Since volume \(V\) and temperature \(T\) are constant for both gases, we will use their pressure ratio \(4:3\) and the molar mass ratio \(2:3\) to find their density ratio: \(\rho_1/\rho_2 = (P_1 * molar mass_1)/(P_2 * molar mass_2) = (4 * 2)/(3 * 3) = 8/9\).

Key concepts

Understanding Partial Pressures

When dealing with mixtures of gases, such as the non-reactive monoatomic ideal gases in our exercise, a fundamental concept is partial pressures. The partial pressure of a gas is the pressure it would exert if it alone occupied the entire volume of the mixture at the same temperature.

Imagine a room filled with balloons of different colors, with each color representing a different gas. The force exerted by the balloons on the walls of the room can be seen as the total pressure. Now, if we imagine removing all balloons but the red ones, the force these red balloons exert is analogous to the partial pressure of that 'red gas'.

In our scenario, the gases have partial pressure ratios of 4:3. Following Dalton’s Law of Partial Pressures, the total pressure in the vessel is the sum of the partial pressures of the individual gases. This knowledge is pivotal as it allows one to determine ratios of the amount of substance, and when combined with other properties, discern insights into molar masses and densities, leading to a deeper understanding of the behavior and characteristics of the gas mixture.

The Role of Molar Mass Ratio

The molar mass of a gas is the mass of one mole of that gas. It's one of the cornerstones in understanding gas behaviour under different conditions, especially when using the ideal gas law. The molar mass ratio provides us with a direct comparison between the weights of moles of different substances.

In the given exercise, the atomic masses of two monoatomic gases are in the ratio 2:3. Since molar mass is directly proportional to the atomic mass for monoatomic gases, we can conclude that the molar masses of the gases have the same ratio. This ratio plays a crucial part in determining the relative densities of the gases when using the ideal gas equation. It's like comparing two sacks of apples and oranges where the ratio tells us that for every two apples, there are three oranges based on weight, allowing us to predict and compare how heavy a specific number of each fruit would be.

Decoding Density Ratio

The density of a substance is an important physical property, indicating how much mass is contained in a given volume. The density ratio is essentially a comparison of how tightly packed the particles of two substances are within the same space.

The density \(\rho\) of a gas can be calculated from the ideal gas law as \(\rho = \frac{mass}{V} = \frac{PVmolar mass}{RT}\). In a comparison of two gases within the same conditions of temperature and volume, we can use their respective partial pressures and molar masses to determine the ratio of their densities.

In our specific example, with the partial pressures and molar mass ratios provided, we calculated that the density ratio is 8:9. This means that when measuring equal volumes of both gases under identical conditions, the mass of the first gas is slightly less than that of the second. It's like comparing the weight of equal volumes of two kinds of stuffing for plush toys; one is slightly lighter, possibly because it's less compact or made of a lighter material.