Problem 27 - A Complete Resource Book in Chemistry for JEE Main

Question

4 grams of an ideal gas occupies \(5.6035\) litres of volume at \(546 \mathrm{~K}\) and 2 atm pressure. What is its molecular weight? (a) 4 (b) 16 (c) 32 (d) 64

Step-by-Step Solution

Answer

The molecular weight of the gas is 16 g/mol (option b).

Step 1: Understand the Ideal Gas Law

The ideal gas law is represented by the equation: \( PV = nRT \), where \( P \) is the pressure of the gas, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.

Step 2: Convert Units as Required

Recall that pressure is given in atmospheres and volume in litres, which is compatible with the standard value of the ideal gas constant \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \). We'll use \( P = 2 \, \text{atm} \), \( V = 5.6035 \, \text{L} \), and \( T = 546 \, \text{K} \).

Step 3: Rearrange the Ideal Gas Equation

First, express \( n \) in terms of the other variables using the equation: \( n = \frac{PV}{RT} \).

Step 4: Calculate the Number of Moles

Substitute the known values into the equation: \( n = \frac{2 \, \text{atm} \times 5.6035 \, \text{L}}{0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \times 546 \, \text{K}} \approx 0.25 \text{ moles}.\)

Step 5: Determine Molecular Weight

By definition, molecular weight \( M \) is the weight of one mole. We know 4 grams is equivalent to \( 0.25 \) moles. Therefore, calculate molecular weight: \( M = \frac{4 \, \text{grams}}{0.25 \, \text{moles}} = 16 \text{ grams/mol}. \)

Key Concepts

Molecular Weight Calculation

Molecular weight, also known as molar mass, refers to the mass of a given substance divided by the amount of substance, measured in moles. It's typically expressed in grams per mole (g/mol). Calculating molecular weight is a fundamental aspect of chemistry, particularly when working with gases. It's essential to understand that molecular weight provides a bridge between the macroscopic and microscopic worlds—translating mass into moles, a measure of particles.
To find the molecular weight, follow these steps:

• First, determine the mass of the substance in grams. In our example, the gas has a mass of 4 grams.
• Then, find the number of moles of the substance. In this exercise, we have calculated that there are 0.25 moles of gas present.
• Finally, use the formula:
  \[M = \frac{\text{grams of substance}}{\text{moles of substance}}\]In this case, the molecular weight, \( M \), is \( \frac{4 \text{ grams}}{0.25 \text{ moles}} = 16 \text{ grams/mol} \).

Understanding how to compute molecular weight helps in numerous chemical contexts, including stoichiometric calculations and reactions involving gases.

Moles of Gas

In chemistry, a mole is a fundamental unit used to measure the amount of substance. It allows chemists to count particles in a practical manner. One mole equals Avogadro's number of particles, which is approximately \(6.022 \times 10^{23}\) particles. This concept becomes handy when dealing with gases, as it simplifies the description of large numbers of molecules or atoms.
To determine moles in gases, we often use the ideal gas law equation:
\[PV = nRT\]Where:

• \( P \) is the pressure in atmospheres (atm).
• \( V \) is the volume in liters (L).
• \( n \) is the number of moles.
• \( R \) is the Ideal Gas Constant, \(0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1}\).
• \( T \) is the temperature in Kelvin (K).

By rearranging the equation to solve for \( n \), we have:\[n = \frac{PV}{RT}\]Using the values given in our scenario—where \( P = 2 \text{ atm} \), \( V = 5.6035 \text{ L} \), and \( T = 546 \text{ K} \)—we calculated the number of moles as approximately 0.25 mol. Understanding the concept of moles helps in predicting how gases will behave under various conditions and is vital in gas-related calculations.

Pressure-Volume-Temperature Relationship

The Pressure-Volume-Temperature (PVT) relationship is a crucial concept in understanding how gases behave. According to the ideal gas law, these three properties are interconnected in the equation:
\[PV = nRT\]This relationship tells us that any change in one of the factors (pressure, volume, or temperature) will affect the others, assuming that the amount of gas measured in moles remains constant.

Key Relationships:

• Pressure and Volume: At a constant temperature, as described by Boyle's Law, the pressure of a gas varies inversely with its volume (\(P \sim \frac{1}{V}\)). This means if you increase the volume, the pressure decreases, assuming temperature remains constant.
• Volume and Temperature: At constant pressure, Charles's Law states that the volume of a gas is directly proportional to its temperature (\(V \sim T\)). Thus, increasing the temperature of a gas, while keeping pressure fixed, will cause an increase in volume.
• Pressure and Temperature: Gay-Lussac's Law states that at constant volume, the pressure of a gas is directly proportional to its temperature (\(P \sim T\)). For example, heating a gas in a sealed container will cause the pressure to rise.

Understanding these relationships enables you to predict and calculate the behavior of gases in a variety of situations, aiding in their applications across different fields such as engineering, meteorology, and environmental science.