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

Question

One litre of gas \(\mathrm{A}\) at \(2 \mathrm{~atm}\) pressure at \(27^{\circ} \mathrm{C}\) and two litres of gas \(\mathrm{B}\) at 3 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 is approximately 3.25 atm (option b).

Step 1: Convert Temperatures to Kelvin

First, we need to convert the given Celsius temperatures to Kelvin. For gas \(A\), the temperature is \(27^\circ C\), which converts to \(300\ K\) (\(T = 27 + 273\)). For gas \(B\), the temperature is \(127^\circ C\), which converts to \(400\ K\) (\(T = 127 + 273\)). The temperature of the mixture is \(327^\circ C\), which converts to \(600\ K\) (\(T = 327 + 273\)).

Step 2: Calculate Moles of Gas A

Using the ideal gas law \(PV = nRT\), calculate the moles of gas \(A\): \(P_A = 2\ atm\), \(V_A = 1\ L\), and \(T_A = 300\ K\). Rearranging for \(n_A\), \(n_A = \frac{P_A V_A}{R T_A} = \frac{2 \times 1}{0.0821 \times 300} = 0.0811\ moles\).

Step 3: Calculate Moles of Gas B

Similarly, calculate the moles of gas \(B\): \(P_B = 3\ atm\), \(V_B = 2\ L\), and \(T_B = 400\ K\). Rearranging for \(n_B\), \(n_B = \frac{P_B V_B}{R T_B} = \frac{3 \times 2}{0.0821 \times 400} = 0.1829\ moles\).

Step 4: Calculate Total Moles of Gas

Add the moles of \(A\) and \(B\) to find the total moles in the mixture: \(n_{total} = n_A + n_B = 0.0811 + 0.1829 = 0.264\ moles\).

Step 5: Use Ideal Gas Law to Find Total Pressure

Using the ideal gas law for the mixture, where \(V_{total} = 4\ L\), and \(T_{mixture} = 600\ K\), solve for total pressure \(P_{total}\): \(P_{total} = \frac{n_{total} R T_{mixture}}{V_{total}} = \frac{0.264 \times 0.0821 \times 600}{4} = 3.25\ atm\).

Step 6: Select the Correct Answer Option

From the calculated total pressure of the mixture \(3.25\ atm\), the correct answer option is (b) \(3.25\ atm\).

Key Concepts

Mole Calculation

The concept of moles is central to understanding how gases behave in the Ideal Gas Law. In chemistry, a mole refers to a quantity of substance that contains as many elementary entities as there are atoms in 12 grams of carbon-12. This number is approximately 6.022 x 10^23 and is known as Avogadro's number.
To find the number of moles of gas in a container, we use the Ideal Gas Law: \[ PV = nRT \]where:

• \(P\) is the pressure of the gas.
• \(V\) is the volume of the gas.
• \(n\) is the number of moles we want to find.
• \(R\) is the ideal gas constant, approximately 0.0821 L·atm/(K·mol).
• \(T\) is the temperature in Kelvin.

Rearranging the equation to find moles: \[ n = \frac{PV}{RT} \]With this formula, you can calculate the moles of each gas before they are mixed. For example, gas A had a pressure of 2 atm, a volume of 1 L, and was at 300 K. Plugging these into the equation yields 0.0811 moles. Gas B, with a pressure of 3 atm, volume of 2 L, and a temperature of 400 K, resulted in 0.1829 moles.

Gas Mixing

When gases are mixed, their individual molecular particles freely intersperse with one another in the volume of the container. This concept is especially helpful in understanding the total behavior of these gases as a mixture.
In mixing gas A and gas B in a container that's 4 liters in size, we need to account for both the volume each gas takes up and the interactions between their molecules. Since we know the number of moles of each gas, we add them together to get the total moles in the container:\[ n_{total} = n_A + n_B \]This simple summation results because we assume ideal behavior—no particle interaction and gases mix completely to occupy the volume of the container without contraction or expansion. For our problem, the mixture contains 0.264 moles of gas in total, which is the sum of moles of gases A and B.

Temperature Conversion

Temperature is a crucial factor when dealing with gas laws. For calculations, temperatures must be in Kelvin to maintain consistency and accuracy. This scale is absolute, starting at absolute zero, the point at which particles have minimum thermal motion.
To convert Celsius to Kelvin, which is often needed in Ideal Gas Law problems, you add 273 to the Celsius temperature. Here are the necessary conversions from the exercise:

• Gas A: From 27°C to 300 K
• Gas B: From 127°C to 400 K
• Mixture temperature: From 327°C to 600 K

By performing this simple conversion, you ensure that your calculations align with the requirements of the Ideal Gas Law. Remember, accurate temperature conversion is a necessary step before using any gas equation for calculations.

Pressure Calculation

One of the significant goals in studying gas systems is determining the pressure of a mixture. After understanding the amounts and temperatures of involved gases, we use the Ideal Gas Law to find the total pressure of the gas mixture.
The formula rearranged for pressure is:\[ P = \frac{nRT}{V} \]Here, you use the total moles of the gas mixture, the temperature of the mixed gases in Kelvin, and the volume of the container. In our case, the total moles were 0.264, the temperature was 600 K, and the volume was 4 liters.
Plug these values into the equation:\[ P_{total} = \frac{0.264 imes 0.0821 imes 600}{4} = 3.25\, \text{atm} \]This pressure provides insights into the behavior of the gas mixture under the given conditions.