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If equal amounts of helium and neon are placed in a porous container, which gas will escape faster?

Short Answer

Expert verified
Helium will escape from the porous container faster than Neon.

Step by step solution

01

Understand Graham's Law of Diffusion

Graham's Law of Diffusion states that the rate of diffusion (or effusion) of a gas is inversely proportional to the square root of its molar mass. In this case, we're looking at the difference in diffusion rates between helium and neon.
02

Determine the molar mass of the gases

Helium (He) has a molar mass of about 4 g/mol, while Neon (Ne) has a molar mass of about 20 g/mol.
03

Apply Graham's Law of Diffusion

As per Graham's Law, the rate of diffusion of a gas is faster when it has a lower molar mass. Since helium has a lower molar mass than neon, it will escape the porous container more quickly.

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Key Concepts

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

Gas Effusion Rate
The effusion rate of a gas is a measure of how quickly it escapes through a tiny opening into a vacuum. Understanding the effusion rate is critical because it can determine how we use gases in practical applications, such as in the distribution of gases through pipelines or the inflating of airships.

To determine which gas will escape faster, helium or neon, we must consider Graham's Law of Effusion. This law states that the effusion rate of a gas is inversely proportional to the square root of its molar mass, which is represented mathematically as: \[ R_1 / R_2 = \sqrt{M_2 / M_1} \] where \( R_1 \) and \( R_2 \) are the effusion rates of gases 1 and 2 respectively, and \( M_1 \) and \( M_2 \) are their respective molar masses.

This relationship tells us that lighter gases (with lower molar mass) effuse more rapidly than heavier gases. For instance, in our example with helium and neon, we would expect helium to effuse faster due to its significantly lower molar mass.
Molar Mass
Molar mass is a fundamental concept in chemistry and is defined as the mass of one mole of a substance. It is typically expressed in units of grams per mole (g/mol). This property is particularly crucial when discussing Graham's Law because the law is grounded in the relationship between the molar mass of a gas and its rate of diffusion or effusion.

Helium, with a molar mass of about 4 g/mol, is much lighter than neon, which has a molar mass of about 20 g/mol. As outlined in Graham's Law, the lighter molar mass of helium signifies a faster effusion rate when compared to neon. By knowing the molar mass of each gas, you can predict not only their rates of effusion but also how they will behave in mixtures, how they will separate under certain conditions, and how they will transfer energy.

When working with gases, it is also crucial to remember that molar mass impacts not only effusion and diffusion rates but also other physical properties, such as density and how the gas interacts with other substances.
Diffusion in Gases
Diffusion in gases pertains to the movement of gas molecules from an area of higher concentration to an area of lower concentration until the gas is uniformly distributed. This process is driven by the natural kinetic energy that gas molecules possess. The rate of this movement depends on several factors, including temperature, pressure, and the molar mass of the gas molecules.

According to Graham's Law of Diffusion, similar to effusion, the rate of gas diffusion is inversely proportional to the square root of its molar mass. Therefore, a gas like helium will not only effuse faster than neon, but it will also diffuse more rapidly in a mixture or when in contact with another gas.

This is particularly important in fields such as environmental science, where the diffusion rates of gases through the atmosphere can impact pollution dispersion, and in industrial processes, where the separation of gases based on their molar masses is required. The concept of diffusion is also fundamental when studying how gases interact within living systems, including how oxygen and carbon dioxide are exchanged in the respiratory process.

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Most popular questions from this chapter

Given that \(P V=n R T\), solve for the unknown quantity. $$ \begin{array}{|l|l|l|l|l|} \hline \multicolumn{1}{|c|}{\boldsymbol{P}} & \multicolumn{1}{c|}{\boldsymbol{V}} & \multicolumn{1}{c|}{n} & \multicolumn{1}{c|}{\boldsymbol{R}} & \multicolumn{1}{c|}{\boldsymbol{T}} \\ \hline 1.00 & 1.00 & 0.500 & 0.08206 & ? \\ \hline 0.750 & 3.00 & ? & 0.08206 & 237 \\ \hline 3.25 & ? & 1.50 & 0.08206 & 455 \\ \hline ? & 15.0 & 2.67 & 0.08206 & 322 \\ \hline \end{array} $$

A series of organic compounds called the alkanes has the general formula \(\mathrm{C}_{n} \mathrm{H}_{2 n+2}\). Plot the boiling point of the alkanes versus the number of carbon atoms in the alkane. Is there a pattern? Are the data linear? $$ \begin{array}{|c|c|} \hline \text { Alkane } & \text { Boiling Point ( }{ }^{\circ} \mathbf{C} \text { ) } \\ \hline \mathrm{CH}_{4} & -161 \\ \hline \mathrm{C}_{2} \mathrm{H}_{6} & -89 \\ \hline \mathrm{C}_{3} \mathrm{H}_{3} & -44 \\ \hline \mathrm{C}_{4} \mathrm{H}_{10} & -0.5 \\ \hline \mathrm{C}_{5} \mathrm{H}_{12} & 36 \\ \hline \mathrm{C}_{6} \mathrm{H}_{14} & 68 \\ \hline \mathrm{C}_{7} \mathrm{H}_{16} & 98 \\ \hline \mathrm{C}_{8} \mathrm{H}_{18} & 126 \\ \hline \mathrm{C}_{9} \mathrm{H}_{20} & 151 \\ \hline \mathrm{C}_{10} \mathrm{H}_{22} & 174 \\ \hline \end{array} $$

Consider a gas in a container that can adjust its volume to maintain constant pressure. Suppose the gas is cooled. What happens to the gas particles with the decrease in temperature? What happens to the volume of the container?

A tank contains \(78.0 \mathrm{~g}\) of \(\mathrm{N}_{2}\) and \(42.0 \mathrm{~g}\) of \(\mathrm{Ne}\) at a total pressure of \(3.75 \mathrm{~atm}\) and a temperature of \(50.0^{\circ} \mathrm{C}\). Calculate the following quantities. (a) moles of \(\mathrm{N}_{2}\) (c) partial pressure of \(\mathrm{N}_{2}\) (b) moles of \(\mathrm{Ne}\) (d) partial pressure of \(\mathrm{Ne}\)

Plot the velocity of sound at \(25^{\circ} \mathrm{C}\) as a function of the density of the liquid medium through which it is passing. Is there a relationship between these variables? What is the value of the velocity for a liquid with a density of \(0.92 \mathrm{~g} / \mathrm{mL}\) ? $$ \begin{array}{|l|c|c|} \hline \text { Liquid } & \begin{array}{l} \text { Velocity of } \\ \text { Sound }(\mathrm{m} / \mathrm{s}) \end{array} & \text { Density }(\mathrm{g} / \mathrm{mL}) \\ \hline \text { ethyl ether } & 985 & 0.713 \\ \hline \text { acetone } & 1174 & 0.79 \\ \hline \text { ethanol } & 1207 & 0.79 \\ \hline \text { methanol } & 1103 & 0.791 \\ \hline \text { kerosene } & 1324 & 0.81 \\ \hline \text { benzene } & 1295 & 0.870 \\ \hline \text { turpentine } & 1255 & 0.88 \\ \hline \text { castor oil } & 1477 & 0.969 \\ \hline \text { distilled water } & 1497 & 0.998 \\ \hline \text { seawater } & 1531 & 1.025 \\ \hline \text { ethylene glycol } & 1658 & 1.113 \\ \hline \text { glycerol } & 1904 & 1.26 \\ \hline \end{array} $$

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