Problem 94 - A Complete Resource Book in Chemistry for JEE Main

Question

One litre of gas \(\mathrm{A}\) at 2 atm pressure at \(27^{\circ} \mathrm{C}\) and two litres of gas \(\mathrm{B}\) at \(3 \mathrm{~atm}\) pressure at \(127^{\circ} \mathrm{C}\) are mixed in a 4 litre vessel. The temperature of the mixture is maintained at \(327^{\circ} \mathrm{C}\). What is the total pressure of the gaseous mixture? (a) \(3.93 \mathrm{~atm}\) (b) \(3.25 \mathrm{~atm}\) (c) \(4.25 \mathrm{~atm}\) (d) \(6.25 \mathrm{~atm}\)

Step-by-Step Solution

Answer

The total pressure of the gaseous mixture is \(3.25 \text{ atm}\).

Step 1: Calculate the moles of Gas A

We start by using the ideal gas law to find the number of moles of Gas A. The formula for the ideal gas law is \( PV = nRT \). For Gas A, we have \( P = 2 \text{ atm} \), \( V = 1 \text{ L} \), and \( T = 27^{\circ} \text{C} = 300 \text{ K} \) (since \( T(K) = T(°C) + 273 \)). Thus, \( n = \frac{PV}{RT} = \frac{2 \times 1}{0.0821 \times 300} \approx 0.081 \text{ moles} \).

Step 2: Calculate the moles of Gas B

Similarly, for Gas B, we use the ideal gas law \( PV = nRT \) with \( P = 3 \text{ atm} \), \( V = 2 \text{ L} \), and \( T = 127^{\circ} \text{C} = 400 \text{ K} \). Hence, \( n = \frac{PV}{RT} = \frac{3 \times 2}{0.0821 \times 400} \approx 0.183 \text{ moles} \).

Step 3: Calculate the total moles of gas

Now, we add the moles of Gas A and Gas B to find the total moles in the mixture: \( n_{\text{total}} = 0.081 + 0.183 = 0.264 \text{ moles} \).

Step 4: Calculate the total pressure of the mixture

We apply the ideal gas law again to find the total pressure of the gaseous mixture. The volume of the vessel is 4 L and the temperature is \(327^{\circ} \text{C} = 600 \text{ K} \). With \( n_{\text{total}} = 0.264 \text{ moles} \), the formula gives us: \( P_{\text{total}} = \frac{nRT}{V} = \frac{0.264 \times 0.0821 \times 600}{4} \approx 3.25 \text{ atm} \).

Key Concepts

Calculating Moles

When using the Ideal Gas Law, a crucial step is calculating the number of moles of a gas, denoted by \( n \). This calculation is necessary to understand gas behavior in different conditions of pressure, volume, and temperature. The Ideal Gas Law is represented as \( PV = nRT \). Here, \( P \) stands for pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the gas constant (approximately 0.0821 atm·L/mol·K), and \( T \) is the temperature in Kelvin. This equation helps us establish the relationship between these variables.

To calculate the moles of a gas:

• Rearrange the formula to solve for \( n \): \( n = \frac{PV}{RT} \).
• Plug in the values for pressure, volume, and temperature (converted to Kelvin, remember!).
• Divide the product of pressure and volume by the product of the gas constant and temperature to find \( n \).

Let's apply this with the gases from our exercise:
For Gas A, where \( P = 2 \text{ atm} \), \( V = 1 \text{ L} \), and \( T = 300 \text{ K} \) (since \( 27^{\circ} \text{C} \) + 273 = 300 K), the moles are approximately \( n_A = 0.081 \). Similarly, for Gas B with \( P = 3 \text{ atm} \), \( V = 2 \text{ L} \), and \( T = 400 \text{ K} \), the moles are about \( n_B = 0.183 \). The total number of moles in the mixture is \( n_{\text{total}} = 0.264 \).

Total Pressure of Gaseous Mixture

Once the moles are calculated, the next step is determining the total pressure of the gaseous mixture. This step considers all the gases in the mixture, accounting for the overall pressure they exert within the container.

We continue using the Ideal Gas Law \( PV = nRT \). Here, to find the total pressure \( P_{\text{total}} \), we'll need:

• Total number of moles of gas, \( n_{\text{total}} \).
• The volume of the mixture, \( V \).
• The temperature of the mixture, which must be in Kelvin.

The formula can be rearranged to solve for \( P \): \[ P_{\text{total}} = \frac{nRT}{V} \].
Plug in \( n_{\text{total}} = 0.264 \text{ moles} \), \( R = 0.0821 \), \( T = 600 \text{ K} \) (because \( 327^{\circ} \text{C} \) + 273 = 600 K), and \( V = 4 \text{ L} \). Substituting these values, the total pressure comes out to approximately \( 3.25 \text{ atm} \). This value shows how gases, when mixed, reach equilibrium, distributing their pressure across the available volume.

Temperature Conversion from Celsius to Kelvin

Converting Celsius to Kelvin is a vital skill when dealing with gases, as the Ideal Gas Law requires temperature inputs in Kelvin. This conversion ensures accurate calculations since Kelvin is the absolute temperature scale used in scientific equations.

The conversion between Celsius and Kelvin is straightforward:

• Add 273 to the Celsius temperature to convert it to Kelvin.

For example, let's see how this works for our gases:

• Gas A is initially at \( 27^{\circ} \text{C} \). Convert this by adding 273. Resulting in a temperature of \( 300 \text{ K} \).
• Gas B starts at \( 127^{\circ} \text{C} \). Add 273 to get \( 400 \text{ K} \).
• The mixture temperature is given as \( 327^{\circ} \text{C} \), which becomes \( 600 \text{ K} \) when converted.

This conversion is crucial because it aligns the temperature with the Kelvin scale, which directly correlates with the internal energy of gas particles. It ensures consistency and accuracy across calculations involving gases.