Question

If 0.214 mole of argon gas occupies a volume of \(652 \mathrm{~mL}\) at a particular temperature and pressure, what volume would 0.375 mole of argon occupy under the same conditions?

Short answer

0.375 mol of argon would occupy approximately 1143 mL volume under the same conditions as the initial moles of argon. This is calculated using the direct proportion formula, \(V_1/N_1 = V_2/N_2\), where V₁ and N₁ are the initial volume and moles, and V₂ and N₂ are the final volume and moles.

Step-by-step solution

Identify relevant information

We are given: - Initial moles of argon: 0.214 mol - Initial volume: 652 mL - Final moles of argon: 0.375 mol - Final volume: unknown (V₂) The pressure and temperature are constant.

Use the direct proportion formula

Since pressure and temperature are constant, we can use the direct proportion formula: \(V_1/N_1 = V_2/N_2\) where V₁ and N₁ are the initial volume and moles, and V₂ and N₂ are the final volume and moles.

Plug in given values

Now, substitute the given values into the formula: \(652 \mathrm{~mL} / 0.214 \textrm{ mol} = V_2 / 0.375 \textrm{ mol}\)

Solve for the final volume

To find the final volume V₂, multiply both sides of the equation by 0.375 mol: \(V_2 = \frac{(652 \mathrm{~mL})(0.375 \textrm{ mol})}{0.214 \textrm{ mol}}\)

Calculate the final answer

Multiply and divide the numbers to find V₂: \(V_2 = \frac{(652 \mathrm{~mL})(0.375 \textrm{ mol})}{0.214 \textrm{ mol}} \approx 1142.99 \mathrm{~mL}\) So, 0.375 mol of argon would occupy approximately 1143 mL volume under the same conditions.

Key concepts

Mole Concept

The mole is a fundamental unit in chemistry that helps in quantifying the amount of a substance. One mole is equivalent to Avogadro's number, which is approximately \(6.022 \times 10^{23}\) particles, whether they be atoms, molecules, ions, or electrons. This concept allows chemists to count entities at the microscopic level using macroscopic measurements.

• Definition: A mole is 1 mole of any substance holds exactly the number of units equivalent to Avogadro's number.
• Relevance: It enables conversion between the microscopic world (individual particles) and the macroscopic world (grams, liters).

In our problem, the concept of mole is used to relate the number of moles of argon gas to the volume it occupies. Since gases behave in predictable ways under standard conditions, knowing how many moles of gas you have lets you predict other properties, such as volume.

Volume Calculation

Volume calculation in chemistry often involves gases, which typically occupy the container they are in. Different gases at identical temperature and pressure occupy the same volume due to a principle known as Avogadro's Law, which states that equal volumes of all gases, at the same temperature and pressure, have the same number of molecules.In the given exercise, to find the volume of argon gas when its amount changes from 0.214 moles to 0.375 moles but the conditions stay constant, we use a simple volume calculation. The formula used is a proportion based on the known volume for a given amount of substance:- Start with the volume given initially (652 mL for 0.214 mol)- Apply the proportion formula to find the new volume: \[ V_2 = \frac{V_1 \times N_2}{N_1} \]Where:- \( V_1 \) is the initial volume, 652 mL.- \( N_1 \) is the initial moles, 0.214 mol.- \( N_2 \) is the final moles, 0.375 mol.

Direct Proportion in Chemistry

Direct proportion is a relationship between two quantities in which they increase or decrease at the same rate. When one quantity doubles, so does the other. In the context of gas laws, under constant temperature and pressure, the volume of a gas is directly proportional to the number of moles.

• Definition: Two quantities are directly proportional if, by multiplying one by a factor, the other is multiplied by the same factor.
• Examples: Both average speed and distance in motion equations display direct relationships.

In our exercise, the problem revolves around the direct proportion of volume to the number of moles of argon gas. The formula \( \frac{V_1}{N_1} = \frac{V_2}{N_2} \) makes it straightforward to compute unknowns when given such a relationship in chemistry and is essential when dealing with ideal gases in fixed conditions. Thus, if the moles increased by a given ratio, the volume must also increase by the same ratio, provided that the pressure and temperature remain unchanged.