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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?

Short Answer

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

Step by step solution

01

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: \(n_2\)
02

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: \( n_2 = \frac{0.50 mol}{11.1 L} \cdot 20 L \)
03

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Small quantities of hydrogen gas can be prepared in the laboratory by the addition of aqueous hydrochloric acid to metallic zinc. $$\mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)$$ Typically, the hydrogen gas is bubbled through water for collection and becomes saturated with water vapor. Suppose 240\. mL of hydrogen gas is collected at \(30 .^{\circ} \mathrm{C}\) and has a total pressure of \(1.032\) atm by this process. What is the partial pressure of hydrogen gas in the sample? How many grams of zinc must have reacted to produce this quantity of hydrogen? (The vapor pressure of water is 32 torr at \(30^{\circ} \mathrm{C}\).)

You have a sealed, flexible balloon filled with argon gas. The atmospheric pressure is \(1.00\) atm and the temperature is \(25^{\circ} \mathrm{C}\). Assume that air has a mole fraction of nitrogen of \(0.790\), the rest being oxygen. a. Explain why the balloon would float when heated. Make sure to discuss which factors change and which remain constant, and why this matters. b. Above what temperature would you heat the balloon so that it would float?

We state that the ideal gas law tends to hold best at low pressures and high temperatures. Show how the van der Waals equation simplifies to the ideal gas law under these conditions.

A person accidentally swallows a drop of liquid oxygen, \(\mathbf{O}_{2}(l)\) which has a density of \(1.149 \mathrm{g} / \mathrm{mL}\). Assuming the drop has a volume of \(0.050 \mathrm{mL},\) what volume of gas will be produced in the person's stomach at body temperature \(\left(37^{\circ} \mathrm{C}\right)\) and a pressure of \(1.0 \mathrm{atm}\)?

Consider two gases, \(A\) and \(B\), each in a \(1.0\) -\(\mathrm{L}\) container with both gases at the same temperature and pressure. The mass of gas \(A\) in the container is \(0.34\) \(\mathrm{g}\) and the mass of gas \(B\) in the container is \(0.48 \mathrm{g}\). a. Which gas sample has the most molecules present? Explain. b. Which gas sample has the largest average kinetic energy? Explain. c. Which gas sample has the fastest average velocity? Explain. d. How can the pressure in the two containers be equal to each other since the larger gas \(B\) molecules collide with the container walls more forcefully?

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