Understanding the Total Pressure in a Gas Mixture: A Practical Example

In this article, we will explore the process of determining the total pressure of a gas mixture using the Ideal Gas Law. We will walk through a practical example where additional 80.0 grams of Argon (Ar) and 142 grams of Oxygen (O2) are placed in a 30.0-L gas tank at a temperature of 29 °C.

Given Data and Initial Conversion

Let's start by examining the given data and the necessary conversions to find the moles of each gas:

Molar mass of: Argon (Ar) 39.948 g/molO2 (Oxygen) 2 x 15.999 g/mol 31.998 g/mol Mass given: 80.0 g of Ar142 g of O2

To convert the mass to moles, we use the formula: moles mass / molar mass.

Moles of Ar: 80.0 g / 39.948 g/mol 2.00 mol Moles of O2: 142 g / 31.998 g/mol 4.44 mol

Both Ar and O2 can be treated as ideal gases under the given conditions, making the Ideal Gas Law applicable:

PV nRT, where P is pressure, V is volume, n is moles, R is the gas constant, and T is temperature in Kelvin.

Calculating the Total Pressure

Given: V 30.0 Ln 2.00 mol (Ar) 4.44 mol (O2) 6.44 molR 0.082057 (mol.K)T 29 °C 273.15 302.15 K

To find the total pressure (P), we rearrange the Ideal Gas Law equation to solve for P:

P (nRT) / V

Substituting the values:

P (6.44 mol x 0.082057 (mol.K) x 302.15 K) / 30.0 L

P ≈ 5.32 atm

Conclusion and Key Points

The total pressure of the gas mixture in the 30.0-L tank at 29 °C is approximately 5.32 atm. This example demonstrates the practical application of the Ideal Gas Law to determine the pressure of a gas mixture with known components and conditions.

Key Points:

Ideal Gas Law (PV nRT) Calculation of moles from given mass and molar mass Determination of total pressure in a gas mixture

Understanding these concepts is crucial for various practical applications, including chemical engineering, environmental science, and more.