Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 12

If you have any two gases in different containers that are the same size at the same pressure and same temperature, what is true about the moles of each gas? Why is this true?

Short Answer

Expert verified
The moles of each gas in the two containers are equal. This is true because, according to the Ideal Gas Law, the relationship between the pressure, volume, temperature, and number of moles must remain constant for two gases given the same conditions. Since both containers share the same pressure, volume, and temperature, the equation \(PV = nRT\) is satisfied, and thus the moles of each gas are equal.

Step by step solution

01

Recognize the given information

Given two containers that are the same size, both containers have the same volume (V). Additionally, both gases have the same pressure (P) and temperature (T).
02

Use the Ideal Gas Law

The Ideal Gas Law is given by the equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, T is the temperature, and R is the gas constant. By substituting the values given in the problem, we can find the moles of each gas.
03

Compare the two gases

Let the two containers be labeled A and B. Apply the Ideal Gas Law equation to both containers: Container A: \(P_A V_A = n_A R T_A\) Container B: \(P_B V_B = n_B R T_B\) Since both containers have the same pressure, volume, and temperature, we can write: \(P_A = P_B\) \(V_A = V_B\) \(T_A = T_B\)
04

Equate the Ideal Gas Law equations

We can set the two Ideal Gas Law equations equal to each other, since the pressure, volume, and temperature are the same for both containers: \(P_A V_A = n_A R T_A\) \(P_B V_B = n_B R T_B\) \(PV = n_A R T\) \(PV = n_B R T\)
05

Solve for the relationship between moles of each gas

Dividing the two equations, we find the relationship between the moles of each gas: \(\frac{n_A R T}{n_B R T} = \frac{PV}{PV}\) Canceling out the common terms, we are left with: \(\frac{n_A}{n_B} = \frac{PV}{PV}\) \(\frac{n_A}{n_B} = 1\) Thus, the moles of each gas (n_A and n_B) are equal. This is true because, according to the Ideal Gas Law, the relationship between the pressure, volume, temperature, and number of moles must remain constant for two gases given the same conditions.

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.

Start your free trial

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

Silane, SiH, , is the silicon analogue of methane, \(\mathrm{CH}_{4}\) . It is prepared industrially according to the following equations: $$\begin{array}{c}{\mathrm{Si}(s)+3 \mathrm{HCl}(g) \longrightarrow \mathrm{HSiCl}_{3}(l)+\mathrm{H}_{2}(g)} \\ {4 \mathrm{HSiCl}_{3}(l) \longrightarrow \mathrm{SiH}_{4}(g)+3 \mathrm{SiCl}_{4}(l)}\end{array}$$ a. If \(156 \mathrm{mL}\) \(\mathrm{HSiCl}_{3} (d=1.34 \mathrm{g} / \mathrm{mL})\) is isolated when 15.0 \(\mathrm{L}\) \(\mathrm{HCl}\) at 10.0 \(\mathrm{atm}\) and \(35^{\circ} \mathrm{C}\) is used, what is the percent yield of \(\mathrm{HSiCl}_{3} ?\) b. When \(156 \mathrm{HSiCl}_{3}\) is heated, what volume of \(\mathrm{SiH}_{4}\) at 10.0 \(\mathrm{atm}\) and \(35^{\circ} \mathrm{C}\) will be obtained if the percent yield of the reaction is 93.1\(\% ?\)

One of the chemical controversies of the nineteenth century concerned the element beryllium (Be). Berzelius originally claimed that beryllium was a trivalent element (forming \(\mathrm{Be}^{3+}\) ions) and that it gave an oxide with the formula \(\mathrm{Be}_{2} \mathrm{O}_{3}\) . This resulted in a calculated atomic mass of 13.5 for beryllium. In formulating his periodic table, Mendeleev proposed that beryllium was divalent (forming \(\mathrm{Be}^{2+}\) ions) and that it gave an oxide with the formula BeO. This assumption gives an atomic mass of \(9.0 .\) In \(1894,\) A. Combes (Comptes Rendus \(1894,\) p. 1221 ) reacted beryllium with the anion $C_{5} \mathrm{H}_{7} \mathrm{O}_{2}^{-}$ and measured the density of the gaseous product. Combes's data for two different experiments are as follows: $$\begin{array}{lll}{\text { Mass }} & {0.2022 \mathrm{g}} & {0.2224 \mathrm{g}} \\ {\text { Volume }} & {22.6 \mathrm{cm}^{3}} & {26.0 \mathrm{cm}^{3}} \\ {\text { Temperature }} & {13^{\circ} \mathrm{C}} & {17^{\circ} \mathrm{C}} \\ {\text { Pressure }} & {765.2 \mathrm{mm} \mathrm{Hg}} & {764.6 \mathrm{mm}}\end{array}$$ If beryllium is a divalent metal, the molecular formula of the product will be \(\mathrm{Be}\left(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2}\right)_{2} ;\) if it is trivalent, the formula will be $\mathrm{Be}\left(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2}\right)_{3} .$ Show how Combes's data help to confirm that beryllium is a divalent metal.

One way of separating oxygen isotopes is by gaseous diffusion of carbon monoxide. The gaseous diffusion process behaves like an effusion process. Calculate the relative rates of effusion of $^{12} \mathrm{C}^{16} \mathrm{O},^{12} \mathrm{C}^{17} \mathrm{O},\( and \)^{12} \mathrm{C}^{18} \mathrm{O} .$ Name some advantages and disadvantages of separating oxygen isotopes by gaseous diffusion of carbon dioxide instead of carbon monoxide.

Silicon tetrachloride \(\left(\mathrm{SiCl}_{4}\right)\) and trichlorosilane \(\left(\mathrm{SiHCl}_{3}\right)\) are both starting materials for the production of electronics-grade silicon. Calculate the densities of pure \(\mathrm{SiCl}_{4}\) and pure \(\mathrm{SiHCl}_{4}\) vapor at $85^{\circ} \mathrm{C}$ and 635 torr.

Hydrogen azide, \(\mathrm{HN}_{3},\) decomposes on heating by the following unbalanced equation: $$\mathrm{HN}_{3}(g) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{H}_{2}(g)$$ If 3.0 atm of pure \(\mathrm{HN}_{3}(g)\) is decomposed initially, what is the final total pressure in the reaction container? What are the partial pressures of nitrogen and hydrogen gas? Assume the volume and temperature of the reaction container are constant.
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Making Measurements

Read Explanation

Chemical Analysis

Read Explanation

Kinetics

Read Explanation

The Earths Atmosphere

Read Explanation

Nuclear Chemistry

Read Explanation
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