Problem 48 - Chemistry for Allied Health

Question

A \(1.85\) mole sample of helium has a volume of \(2.00 \mathrm{~L}\). Additional helium is added at constant pressure and temperature until the volume is \(3.25 \mathrm{~L}\). What is the total moles of helium present in the sample? What mass of helium was added?

Step-by-Step Solution

Answer

The total moles of helium is 3.00625 mol. The mass of helium added is 4.625 g.

Step 1: Understand the Initial Conditions

We start with a gas sample of helium that has an initial volume of \(2.00\, \text{L}\) and contains \(1.85\, \text{mol}\) of helium.

Step 2: Apply the Ideal Gas Law to the Initial Conditions

Recall the Ideal Gas Law: \(PV = nRT\), which can also be adapted to volume and moles when pressure (\(P\)) and temperature (\(T\)) are constant. Here we assume that \(V_1\) and \(n_1\) are proportional values:\[ \frac{V_1}{n_1} = \frac{V_2}{n_2} \] Substitute the known values: \(V_1 = 2.00\, \text{L}\), \(n_1 = 1.85\, \text{mol}\).

Step 3: Understand the Final Conditions

Helium is added, increasing the volume to \(3.25\, \text{L}\). We need to determine the new total moles \(n_2\) of helium now that the volume \(V_2\) is \(3.25\, \text{L}\).

Step 4: Calculate the Total Moles of Helium

Using the relation \(\frac{V_1}{n_1} = \frac{V_2}{n_2}\) from the Ideal Gas Law, we substitute and rearrange to solve for \(n_2\):\[ n_2 = n_1 \times \frac{V_2}{V_1} = 1.85 \, \text{mol} \times \frac{3.25 \, \text{L}}{2.00 \, \text{L}} = 3.00625 \, \text{mol} \]

Step 5: Calculate the Number of Moles Added

Determine the moles added, \(n_{\text{added}}\), by subtracting the initial moles from the total moles:\[ n_{\text{added}} = n_2 - n_1 = 3.00625 \, \text{mol} - 1.85 \, \text{mol} = 1.15625 \, \text{mol} \]

Step 6: Calculate the Mass of Helium Added

The molar mass of helium is approximately \(4.00 \; \text{g/mol}\). Use this to find the mass of helium added:\[ \text{Mass (g)} = n_{\text{added}} \times \text{Molar Mass} = 1.15625 \, \text{mol} \times 4.00 \, \text{g/mol} = 4.625 \, \text{g} \]

Key Concepts

Moles of Gas

The concept of moles is central to understanding chemistry. Moles provide a way to quantify the amount of substance. When dealing with gases, moles offer a counting unit just like dozens or pairs, but for atoms or molecules. Avogadro's number, which is approximately \(6.022 \times 10^{23}\), tells us how many molecules are in one mole of any substance.In the case of gases, moles are particularly useful because they provide a direct relationship with volume, thanks to the Ideal Gas Law. When you know the number of moles of a gas, you can determine how much space it will occupy under specified conditions of pressure and temperature.Understanding moles helps in converting between mass and volume, making problem-solving in chemistry more intuitive. Ultimately, the mole bridges the atomic world with the macroscopic world, helping us predict how gases will behave in different scenarios.

Molar Mass

Molar mass is the mass of one mole of a substance. It is expressed in grams per mole (g/mol) and acts as a bridge between the atomic scale and the real world, letting us convert between the mass of a chemical substance and the quantity in moles.

• For helium, the molar mass is approximately 4.00 g/mol.
• This means one mole of helium weighs exactly 4 grams.

Molar mass becomes especially useful when calculating how much of a substance is required or produced in a chemical reaction. In our scenario, after finding the additional moles of helium added, we multiplied them by helium's molar mass to find the mass added. Thus, understanding molar mass is critical for conversions between the amount of substance in moles and its mass in grams.

Gas Volume

Volume is a key concept when dealing with gases. It is directly related to the amount of gas present when pressure and temperature are constant. At constant conditions, volume changes proportionally with the number of moles present, as demonstrated by the Ideal Gas Law.For our problem:

• Initial volume = 2.00 L associated with 1.85 moles of helium.
• Final volume = 3.25 L after adding more helium.

This proportional relationship means that when additional helium is added and the volume increases to 3.25 L, we use the formula \(n_2 = n_1 \times \frac{V_2}{V_1}\) to find out the total moles at the new volume. It exemplifies how volume changes can determine changes in gas numbers through direct calculation. Mapping these changes through simple mathematics provides us with comprehensive answers about the gas's behavior.

Problem-Solving in Chemistry

Problem-solving in chemistry often involves understanding and manipulating formulas and equations to deduce unknown quantities. It requires a logical approach, combined with an understanding of the key concepts of chemistry like the Ideal Gas Law.In the exercise at hand, we identified:- The initial conditions (volume and moles).- Applied the principle that under constant pressure and temperature, volume and moles are directly proportional.Here, it’s crucial to comprehend that the Ideal Gas Law simplifies to \(\frac{V_1}{n_1} = \frac{V_2}{n_2}\) when temperature and pressure don't change. This allows us to solve for the unknowns using arithmetic calculations. Lastly, converting from the number of moles added to mass using the molar mass demonstrates the linkage between quantitative and qualitative aspects of chemistry. Problem-solving in chemistry, therefore, is as much about applying basic principles as it is about understanding the intricacies of these principles in solving real-world problems.