Problem 49 - A Complete Resource Book in Chemistry for JEE Main

Question

The density of gas \(\mathrm{A}\) is twice of that of \(\mathrm{B}\) and mol, wt. of \(\mathrm{A}\) is half of that of \(\mathrm{B}\). The Ratio of partial pressures of \(\mathrm{P}_{\mathrm{A}}\) and \(\mathrm{P}_{\mathrm{B}}\) is (a) \(\frac{1}{4}\) (b) \(\frac{4}{1}\) (c) \(\frac{2}{1}\) (d) \(\frac{1}{2}\)

Step-by-Step Solution

Answer

The ratio of partial pressures \( P_A : P_B \) is 4:1, option (b).

Step 1: Understand the Given Relations

We are given that the density of gas \(A\) is twice that of gas \(B\), so \( \rho_A = 2 \rho_B \). Also, the molecular weight of \(A\) is half of that of \(B\), so \( M_A = \frac{1}{2} M_B \).

Step 2: Use the Ideal Gas Law

The ideal gas law equation is \( PV = nRT \). To relate density to pressure, we note that \( n = \frac{m}{M} \) where \( m \) is mass and \( M \) is molar mass, and \( \rho = \frac{m}{V} \). Substituting \( n = \frac{\rho V}{M} \) into the ideal gas equation gives \( P = \frac{\rho RT}{M} \).

Step 3: Express Partial Pressures in Terms of Density and Molecular Weight

Using the relation from Step 2, \( P_A = \frac{ \rho_A RT }{ M_A } \) and \( P_B = \frac{ \rho_B RT }{ M_B } \).

Step 4: Substitute Known Relations into the Pressure Ratio

Substitute \( \rho_A = 2 \rho_B \) and \( M_A = \frac{1}{2} M_B \) into the expressions for \( P_A \) and \( P_B \) to find:\[ \frac{P_A}{P_B} = \frac{ \frac{2 \rho_B RT}{\frac{1}{2} M_B} }{ \frac{\rho_B RT}{M_B} } \].

Step 5: Simplify the Expression for the Pressure Ratio

Simplifying the expression from Step 4:\[ \frac{P_A}{P_B} = \frac{2 \cdot \frac{2 \rho_B RT}{M_B}}{\frac{\rho_B RT}{M_B}} = 4 \].Thus, the ratio of partial pressures \( P_A : P_B \) is 4:1.

Key Concepts

Density

Density is a measure of how much mass is contained in a given volume. It is an essential topic in the study of gases and is represented as \( \rho = \frac{m}{V} \). In this formula, \( m \) is the mass of the gas, while \( V \) is the volume it occupies. This relationship helps us to understand how gases with different densities will behave under various conditions. Understanding density is important because we often compare gases by their densities to predict how they will react or mix. In the exercise, the density of gas A was provided as twice that of gas B \( (\rho_A = 2 \rho_B) \). This information becomes pivotal when calculating their individual behaviors using the ideal gas law.

Molecular Weight

The molecular weight (often called molar mass) of a gas is the mass of one mole of its molecules. It plays a crucial role in gas calculations, especially when using the ideal gas law. The molecular weight is denoted as \( M \), commonly expressed in grams per mole. A gas with higher molecular weight will generally have a greater mass for a given volume, which ties back into its density. In our exercise, molecular weight of gas A is stated to be half that of gas B \( (M_A = \frac{1}{2}M_B) \). This reveals how even with differing densities, gases can have significant variance in molecular composition and behavior. Molecular weight is essential in determining how many molecules are present in a sample of gas, which affects everything from its reactivity to its pressure when restrained in a container.

Ideal Gas Law

The ideal gas law is one of the cornerstone equations in chemistry and physics. It relates the pressure, volume, and temperature of a gas to the number of moles present. The equation is expressed as \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is temperature in Kelvin. To understand gases' behavior, it is essential to reorganize the ideal gas equation by incorporating density \( n = \frac{m}{M} = \frac{\rho V}{M} \), allowing us to express pressure as \( P = \frac{\rho RT}{M} \). In this form, it's easier to compare gases and their properties. Applications of the ideal gas law are found everywhere in chemistry, from understanding how gases expand when heated to calculating the amount of gas needed to inflate a balloon to a certain pressure.

Partial Pressure

Partial pressure refers to the pressure exerted by an individual gas within a mixture of gases. According to Dalton's law of partial pressures, the total pressure of a gas mixture is equal to the sum of the partial pressures of the individual gases. Therefore, if you have a mix of gases A and B, you calculate their partial pressures as \( P_A \) and \( P_B \). Each gas contributes to the total pressure based on its fraction of the total volume at a given temperature. When using the ideal gas law for each individual gas, partial pressures can be expressed as \( P_A = \frac{ \rho_A RT }{ M_A } \) and \( P_B = \frac{ \rho_B RT }{ M_B } \). In our exercise, understanding the partial pressure ratio helps us determine the proportionate contributions of gases A and B to the total pressure. Simplifying these expressions informed our solution of the partial pressure ratio, showing how critically each gas behaves in a shared environment.