Problem 45 - Chemistry An Atoms First Approach

Question

An \(11.1\) - \(\mathrm{L}\) sample? sample of gas is determined to contain \(0.50 \) \(\mathrm{mole}\) sample?of \(\mathrm{N}_{2} .\) At the same temperature and pressure, how many moles of gas would there be in a \(20 .\) - \(\mathrm{L}\) sample?

Step-by-Step Solution

Answer

In the 20L container, there would be approximately \(0.899\) moles of gas under the same temperature and pressure conditions.

Step 1: Write down the Ideal Gas Law and the given information.

The ideal gas law is given by the equation: PV = nRT Where, P = Pressure, V = Volume, n = moles of gas, R = Ideal gas constant, and T = Temperature We don't have values for P and T, but since we know P and T remain constant for both samples, we can use the ratio form, given as: \( \frac{n_1}{V_1} = \frac{n_2}{V_2} \) Given: \(V_1 = 11.1 L\), \(n_1 = 0.50 mol\), \(V_2 = 20 L\) We need to find the moles of gas in the 20L container: \(n_2\)

Step 2: Calculate the number of moles in the 20L container using the ratio form of the Ideal Gas Law.

We have the ratio form equation as: \( \frac{n_1}{V_1} = \frac{n_2}{V_2} \) We can solve for \(n_2\) by multiplying both sides by \(V_2\): \( n_2 = \frac{n_1}{V_1} \cdot V_2 \) Now we can substitute the given values into the equation to find the moles of gas in the 20L container: \( n_2 = \frac{0.50 mol}{11.1 L} \cdot 20 L \)

Step 3: Solve for the number of moles in the 20L container.

Calculating the number of moles using the given data: \( n_2 = \frac{0.50 mol}{11.1 L} \cdot 20 L =0.04495495495 mol \cdot 20 L = 0.899099099 mol\) So, in the 20L container, there would be approximately 0.899 moles of gas under the same temperature and pressure conditions.

Key Concepts

Moles of gas

Understanding the concept of moles is fundamental when discussing gases in chemistry. A mole is a unit that represents a specific number of particles, most often atoms or molecules. One mole is equivalent to Avogadro's number, which is approximately \( 6.022 \times 10^{23} \) particles.
The term is essential when dealing with gases because it helps quantify the number of molecules involved without getting lost in extremely large numbers. In a given gas sample, the moles of gas (\( n \)) signify how many moles of molecules are present, which plays a crucial role in calculations involving the Ideal Gas Law.
In the given exercise, the moles of gas in a sample were calculated using the ratio form of the Ideal Gas Law, showcasing how the moles in one volume translate to moles in another volume under constant temperature and pressure.

Volume and pressure relationship

The relationship between volume and pressure in gases is described by Boyle's Law, a principle that shows how, at a constant temperature, the volume of a gas is inversely proportional to its pressure. This means that if the volume of a gas increases, then its pressure decreases, and vice versa, assuming temperature remains unchanged.
However, in the context of the Ideal Gas Law, we often look at scenarios where either volume or pressure varies while maintaining conditions such as temperature. In this exercise, we focused on changing volumes while keeping pressure and temperature constant, which is perfectly captured by the ratio form of the Ideal Gas Law: \( \frac{n_1}{V_1} = \frac{n_2}{V_2} \).
This demonstrates that when volume changes (increases or decreases), the number of moles of gas involved also changes, provided that pressure and temperature don't deviate.

Gas constant

The gas constant \( R \) is a key component of the Ideal Gas Law equation \( PV = nRT \). It serves as a bridge that connects the physical properties of a gas—pressure (\( P \)), volume (\( V \)), moles (\( n \)), and temperature (\( T \)).
\( R \) allows us to use the same equation across different conditions as long as the units used are compatible. For instance, the value of \( R \) is typically \( 8.314 \) J/(mol·K) in SI units, which is designed to ensure consistent results regardless of the specific conditions you are dealing with.
In the exercise, while the explicit value of \( R \) wasn't directly used in calculations, its role was crucial in shaping the understanding that amounts (moles) and conditions (volume and temperature) of gases are consistently interrelated under a set framework.