Problem 128 - A Complete Resource Book in Chemistry for JEE Main

Question

Density of two gases of same molecular weight are in the ratio \(1: 3\) and their temperatures are in the ratio \(3: 2 .\) The ratio of respective pressures is (a) \(2: 1\) (b) \(2: 3\) (c) \(3: 2\) (d) \(1: 2\)

Step-by-Step Solution

Answer

The ratio of their pressures is \(1:2\) (option d).

Step 1: Understanding the Problem

We are given the ratio of densities of two gases as \(1:3\) and the ratio of their temperatures as \(3:2\). We need to find the ratio of their pressures.

Step 2: Using Ideal Gas Law

The Ideal Gas Law states that \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is temperature. For a given mass of gas, \(n\) is proportional to \(dV\), where \(d\) is density. Thus, \(P = \frac{dRT}{M} \), where \(M\) is molecular weight.

Step 3: Setting Up the Ratio Equation

Since the molecular weights are constant, the pressure is directly proportional to the product of density and temperature, i.e., \(P \propto dT\). Therefore, the ratio of the pressures of the two gases is given by \(\frac{P_1}{P_2} = \frac{d_1 T_1}{d_2 T_2}\).

Step 4: Substituting Given Ratios

We know \(\frac{d_1}{d_2} = \frac{1}{3}\) and \(\frac{T_1}{T_2} = \frac{3}{2}\). Substituting these into the ratio equation, we get:\[\frac{P_1}{P_2} = \left( \frac{1}{3} \right) \times \left( \frac{3}{2} \right) = \frac{1}{2}\]

Step 5: Determining the Final Ratio

Thus, the ratio of the pressures \(\frac{P_1}{P_2}\) is \(\frac{1}{2}\), which means \(P_1 : P_2 = 1 : 2\). Hence, the correct answer is option (d) \(1:2\).

Key Concepts

Pressure Calculation

The concept of pressure calculation is crucial when working with gases. Pressure, represented by the symbol \(P\), indicates the force that the gas exerts on the walls of its container. When dealing with gases, we often use the Ideal Gas Law to understand how different factors like volume and temperature affect pressure.

The Ideal Gas Law formula is:

• \[PV = nRT\]

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. In this problem, our focus is on pressure, density, and temperature. We can express pressure in terms of density \(d\) as:

• \[P = \frac{dRT}{M}\]

This equation tells us that pressure is directly proportional to both density and temperature when the molecular weight \(M\) is constant. Thus, understanding how to calculate pressure is key to solving problems involving gaseous substances.

Density Ratio

Density is an important property of gases, representing their mass per unit volume. When comparing two gases, the density ratio gives insight into how dense one gas is compared to another.

In our problem, the ratio of the densities is given as \(1:3\). This means the first gas is less dense than the second. Density is directly tied to pressure and temperature in gases. Using the equation:

• \[P = \frac{dRT}{M}\]

we can see that changes in density alter the pressure when temperature stays the same. In pressure calculations, this density ratio is pivotal. When densities are compared in a ratio form, we can substitute these values into the Ideal Gas Law-derived equation for pressure to find new gas qualities like pressure.

Temperature Ratio

Temperature is a measure of the average kinetic energy of gas molecules. In this exercise, the temperature ratio between the two gases is \(3:2\). This ratio indicates that the first gas is at a higher temperature than the second gas.

Since temperature directly affects pressure in gases, the temperature ratio is an essential factor in pressure calculations. Given that:

• \[P = \frac{dRT}{M}\]

The direct proportionality between pressure and temperature makes this ratio critical. When we know the ratio of temperatures between two gases, we can use it to determine how pressure will vary between those gases, assuming density and molecular weight remain constant. Therefore, accurately using the temperature ratio helps establish correct pressure comparisons in such gas-related problems.