An equimolar mixture of \(\mathrm{H}_{2}\) and \(\mathrm{D}_{2}\) effuses through an orifice (small hole) at a certain temperature. Calculate the composition (in mole fractions) of the gases that pass through the orifice. The molar mass of \(\mathrm{D}_{2}\) is \(2.014 \mathrm{~g} / \mathrm{mol}\)

Short Answer

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Proceeding with the calculations mentioned in Step 3, you will get the mole fractions of \(H2\) and \(D2\) in the mixture after effusion.

Step by step solution

01

State Graham's Law for Effusion

Graham's Law for Effusion states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Meaning, \(\frac{Rate_{H2}}{Rate_{D2}} = \sqrt{\frac{Molar Mass_{D2}}{Molar Mass_{H2}}}\)
02

Apply Graham's Law

As stated, the molar mass of \(\mathrm{D}_{2}\) is \(2.014 \mathrm{~g} / \mathrm{mol}\) and for \(\mathrm{H}_{2}\) it is \(2\times1.008 = 2.016 \mathrm{~g} / \mathrm{mol}\). Substituting these values into the formula mentioned in step 1, we can get the ratio of rates: \(\frac{Rate_{H2}}{Rate_{D2}} = \sqrt{\frac{2.014}{2.016}}\)
03

Find the Mole Fractions

Once we have the ratio of rates, we know that the total rate is the sum of \(Rate_{H2}\) and \(Rate_{D2}\), and the mole fraction of each gas is its effusion rate divided by the total rate. Therefore, Mole fraction of \(H2 = \frac{Rate_{H2}}{Rate_{H2} + Rate_{D2}}\) and Mole fraction of \(D2 = \frac{Rate_{D2}}{Rate_{H2} + Rate_{D2}}\). Calculate these values using the ratio obtained in Step 2.

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Key Concepts

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

Graham's Law of Effusion
Understanding Graham's Law of Effusion is fundamental to grasping how gas effusion occurs. This law tells us that lighter gases (those with a lower molar mass) will effuse faster than heavier gases. In mathematical terms, it's expressed as a ratio: the rate of effusion of one gas compared to another is inversely proportional to the square root of their respective molar masses. An illustrative example can be seen in the exercise, comparing the effusion of hydrogen with deuterium. This relationship implies that gases with smaller molar masses will have higher effusion rates, simply because their smaller mass allows them to move more quickly as per the principles of kinetic energy.
Effusion Rate
The effusion rate of a gas describes how quickly it escapes through a small hole into a vacuum. Quantifying the effusion rate is key to solving problems related to gas mixtures and their behavior under certain conditions. For instance, in the textbook exercise, by applying Graham's Law, we calculate the ratio of effusion rates for \(\mathrm{H}_2\) and \(\mathrm{D}_2\), considering their nearly identical molar masses. The ratio of effusion rates can then be used to determine the composition of the gas mixture after effusion, which is critical in industries where gas purity is essential.
Molar Mass
Molar mass, the mass of one mole of a substance, is pivotal in determining the rate at which different gases effuse. Since gases with a smaller molar mass effuse more rapidly according to Graham's Law, knowing the precise molar mass of each gas in a mixture allows us to predict and calculate the rates at which they will escape through an orifice. This is particularly useful in processes such as isotope separation or in analyzing the composition of atmospheric gases.
Mole Fraction
The mole fraction represents the proportion of a component within a mixture. It is dimensionless and is calculated by dividing the number of moles of the component by the total number of moles in the mixture. In our example, knowing the effusion rates of \(\mathrm{H}_2\) and \(\mathrm{D}_2\), we infer their mole fractions post-effusion, which helps in understanding the new composition of the gas mixture. This concept is particularly important in chemical engineering and process design, determining how mixtures of gases will behave and how they can be manipulated for desired outcomes.
Kinetic Molecular Theory
The Kinetic Molecular Theory provides the underpinnings for Graham's Law, highlighting that gas particles are in constant, random motion, colliding elastically with one another. Key postulates of this theory include the idea that the temperature of a gas is proportional to the average kinetic energy of its particles and that the size of the particles is negligibly small compared to the distances between them. This theory helps us understand why the mass of the particles affects their effusion rates—lighter particles achieve higher velocities at a given temperature, thus effusing more rapidly. When applied to practical scenarios, such as our example problem, this theory not only explains the nature of gas effusion but also aids in predicting the behavior of gases under various conditions.

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Most popular questions from this chapter

(a) A real gas is introduced into a flask of volume \(V\). Is the corrected volume of the gas greater or less than \(V ?\) (b) Ammonia has a larger \(a\) value than neon does (see Table 5.4 ). What can you conclude about the relative strength of the attractive forces between molecules of ammonia and between atoms of neon?

If \(10.00 \mathrm{~g}\) of water are introduced into an evacuated flask of volume \(2.500 \mathrm{~L}\) at \(65^{\circ} \mathrm{C},\) calculate the mass of water vaporized. (Hint: Assume that the volume of the remaining liquid water is negligible; the vapor pressure of water at \(65^{\circ} \mathrm{C}\) is \(187.5 \mathrm{mmHg} .\)

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Air entering the lungs ends up in tiny sacs called alveoli. It is from the alveoli that oxygen diffuses into the blood. The average radius of the alveoli is \(0.0050 \mathrm{~cm}\) and the air inside contains 14 percent oxygen. Assuming that the pressure in the alveoli is 1.0 atm and the temperature is \(37^{\circ} \mathrm{C},\) calculate the number of oxygen molecules in one of the alveoli.

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