Consider a mixture of two gases, \(\mathrm{A}\) and \(\mathrm{B}\), confined in a closed vessel. A quantity of a third gas, \(\mathrm{C}\), is added to the same vessel at the same temperature. How does the addition of gas \(\mathrm{C}\) affect the following: (a) the partial pressure of gas \(A,(\mathbf{b})\) the total pressure in the vessel, (c) the mole fraction of gas B?

Short Answer

Expert verified
(a) The partial pressure of gas A remains unchanged. (b) The total pressure in the vessel increases. (c) The mole fraction of gas B decreases.

Step by step solution

01

Understand the concepts of partial pressure and mole fraction

Partial pressure is the pressure exerted by each individual gas in a mixture as if it were occupying the container by itself. Mole fraction is the ratio of the moles of a particular gas to the total moles of the gaseous mixture.
02

Use Ideal Gas Law for the initial system of gases A and B

We can write the Ideal Gas Law for the initial system containing gases A and B as follows: \[P_{initial}V = n_{A}RT + n_{B}RT\] Where: - \(P_{initial}\) is the initial total pressure of the system - \(V\) is the volume of the vessel - \(n_A\) and \(n_B\) are the moles of gases A and B, respectively - \(R\) is the gas constant - \(T\) is the temperature
03

Add gas C and write the Ideal Gas Law for the final system

Now, we add gas C to the system. Let's write the Ideal Gas Law for the final system containing gases A, B, and C: \[P_{final}V = n_{A}RT + n_{B}RT + n_{C}RT\] Where: - \(P_{final}\) is the final total pressure of the system after adding gas C - \(n_C\) is the moles of gas C added to the system
04

Analyze the effect of adding gas C on the partial pressure of gas A

As the addition of gas C does not change the moles of gas A, its partial pressure, which is given by: \[P_{A} = \dfrac{n_{A}RT}{V}\] will not change because the number of moles, volume, gas constant, and temperature remain the same after adding gas C. (a) Effect on partial pressure of gas A: No change.
05

Analyze the effect of adding gas C on the total pressure in the vessel

Based on the comparison of the equation from step 2 and step 3, we can see that the final pressure increases due to the additional moles of gas C. This means the total pressure in the vessel after adding gas C will be greater than the initial total pressure. (b) Effect on the total pressure in the vessel: Increases.
06

Analyze the effect of adding gas C on the mole fraction of gas B

Mole fraction of gas B is given by the formula: \[\chi_{B} = \dfrac{n_{B}}{n_{A} + n_{B} + n_{C}}\] Comparing the mole fraction of gas B before and after the addition of gas C, initially: \[\chi_{B,initial} = \dfrac{n_{B}}{n_{A} + n_{B}}\] After adding gas C: \[\chi_{B,final} = \dfrac{n_{B}}{n_{A} + n_{B} + n_{C}}\] Since the addition of gas C increases the denominator in the mole fraction expression without affecting the numerator, the mole fraction of gas B will decrease after adding gas C. (c) Effect on the mole fraction of gas B: Decreases.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

On a single plot, qualitatively sketch the distribution of molecular speeds for (a) \(\mathrm{Kr}(g)\) at $-50^{\circ} \mathrm{C},(\mathbf{b}) \mathrm{Kr}(g)\( at \)0^{\circ} \mathrm{C}$ (c) \(\mathrm{Ar}(g)\) at \(0^{\circ} \mathrm{C}\). [Section \(\left.10.7\right]\)

Both Jacques Charles and Joseph Louis Guy-Lussac were avid balloonists. In his original flight in 1783 , Jacques Charles used a balloon that contained approximately \(31,150 \mathrm{~L}\) of \(\mathrm{H}_{2}\). He generated the \(\mathrm{H}_{2}\) using the reaction between iron and hydrochloric acid: $$\mathrm{Fe}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{FeCl}_{2}(a q)+\mathrm{H}_{2}(g)$$ How many kilograms of iron were needed to produce this volume of \(\mathrm{H}_{2}\) if the temperature was \(22{ }^{\circ} \mathrm{C} ?\)

Table 10.3 shows that the van der Waals \(b\) parameter has units of $\mathrm{L} / \mathrm{mol}$. This implies that we can calculate the size of atoms or molecules from \(b\). Using the value of \(b\) for \(\mathrm{Xe},\) calculate the radius of a Xe atom and compare it to the value found in Figure \(7.7,\) that is, \(140 \mathrm{pm}\). Recall that the volume of a sphere is $(4 / 3) \pi r^{3}$.

A 6.53 -g sample of a mixture of magnesium carbonate and calcium carbonate is treated with excess hydrochloric acid. The resulting reaction produces $1.72 \mathrm{~L}\( of carbon dioxide gas at \)28^{\circ} \mathrm{C}\( and \)99.06 \mathrm{kPa}$ pressure. (a) Write balanced chemical equations for the reactions that occur between hydrochloric acid and each component of the mixture. (b) Calculate the total number of moles of carbon dioxide that forms from these reactions. (c) Assuming that the reactions are complete, calculate the percentage by mass of magnesium carbonate in the mixture.

A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required \(105 \mathrm{~s}\) for $1.0 \mathrm{~L}$ of the gas to effuse. Under identical experimental conditions it required \(31 \mathrm{~s}\) for \(1.0 \mathrm{~L}\) of \(\mathrm{O}_{2}\) gas to effuse. Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of $1.0 \mathrm{~L} ;$ in other words, rate is the amount that diffuses over the time it takes to diffuse.)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free