Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 58

A 2.35 L container of \(\mathrm{H}_{2}(\mathrm{g})\) at \(762 \mathrm{mmHg}\) and \(24^{\circ} \mathrm{C}\) is connected to a 3.17 L container of \(\mathrm{He}(\mathrm{g})\) at \(728 \mathrm{mmHg}\) and \(24^{\circ} \mathrm{C}\). After mixing, what is the total gas pressure, in millimeters of mercury, with the temperature remaining at \(24^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The total gas pressure after mixing of the hydrogen and helium gases is the sum of the individual pressures calculated following Boyle's Law.

Step by step solution

01

Convert pressure of H2 gas after mixing

Find the pressure of the hydrogen gas once mixed. According to Boyle's Law, the pressure and volume of a gas have an inverse relationship when temperature is held constant. Hence, \(P_{1}V_{1} = P_{2}V_{2}\). Given \(P_{1} = 762 \, mmHg\) and \(V_{1} = 2.35 \, L\), while \(V_{2}\) is the total volume of the system which is \(V_{1} + V_{2} = 2.35 \, L + 3.17 \, L = 5.52 \, L\). Solve for \(P_{2}\) to get the pressure of H2 gas after mixing.
02

Convert pressure of He gas after mixing

Similar to Step 1, determine the pressure of the helium gas once mixed. With \(P_{1} = 728 \, mmHg\) and \(V_{1} = 3.17 \, L\), and the same \(V_{2} = 5.52 \, L\), we can solve for \(P_{2}\) which will give us the pressure of He gas after mixing.
03

Add the two pressures

The total pressure of the system will be the sum of the pressures of H2 and He after mixing. Calculate the sum of the results obtained in Step 1 and Step 2 to get the total gas pressure of the system.

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

When solid \(\mathrm{KClO}_{3}\) is heated strongly, it decomposes to form solid potassium chloride, \(\mathrm{KCl}\), and \(\mathrm{O}_{2}\) gas. \(\mathrm{A}\) \(0.415 \mathrm{g}\) sample of impure \(\mathrm{KClO}_{3}\) is heated strongly and the \(\mathrm{O}_{2}\) gas produced by the decomposition is collected over water. When the wet \(\mathrm{O}_{2}\) gas is cooled back to \(26^{\circ} \mathrm{C}\), the total volume is \(229 \mathrm{mL}\) and the total pressure is 323 Torr. What is the mass percentage of \(\mathrm{KClO}_{3}\) in the original sample? Assume that none of the impurities produce oxygen on heating. The vapor pressure of water is 25.22 Torr at \(26^{\circ} \mathrm{C}\).

What is the molar volume of an ideal gas at (a) \(25^{\circ} \mathrm{C}\) and 1.00 atm; \((b) 100^{\circ} \mathrm{C}\) and 748 Torr?

A weather balloon filled with He gas has a volume of \(2.00 \times 10^{3} \mathrm{m}^{3}\) at ground level, where the atmospheric pressure is 1.000 atm and the temperature \(27^{\circ}\) C. After the balloon rises high above Earth to a point where the atmospheric pressure is 0.340 atm, its volume increases to \(5.00 \times 10^{3} \mathrm{m}^{3} .\) What is the temperature of the atmosphere at this altitude?

In order for a gas-filled balloon to rise in air, the density of the gas in the balloon must be less than that of air. (a) Consider air to have a molar mass of \(28.96 \mathrm{g} / \mathrm{mol}\) determine the density of air at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{atm},\) in g/L. (b) Show by calculation that a balloon filled with carbon dioxide at \(25^{\circ} \mathrm{C}\) and 1 atm could not be expected to rise in air at \(25^{\circ} \mathrm{C}\)

The equation \(d / P=M / R T,\) which can be derived from equation \((6.14),\) suggests that the ratio of the density \((d)\) to pressure (P) of a gas at constant temperature should be a constant. The gas density data at the end of this question were obtained for \(\mathrm{O}_{2}(\mathrm{g})\) at various pressures at \(273.15 \mathrm{K}\) (a) Calculate values of \(d / P,\) and with a graph or by other means determine the ideal value of the term \(d / P\) for \(\mathrm{O}_{2}(\mathrm{g})\) at \(273.15 \mathrm{K}\) [Hint: The ideal value is that associated with a perfect (ideal) gas.] (b) Use the value of \(d / P\) from part (a) to calculate a precise value for the atomic mass of oxygen, and compare this value with that listed on the inside front cover. $$\begin{array}{lllll} P, \mathrm{mmHg}: & 760.00 & 570.00 & 380.00 & 190.00 \\ d, \mathrm{g} / \mathrm{L}: & 1.428962 & 1.071485 & 0.714154 & 0.356985 \end{array}$$
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

The Earths Atmosphere

Read Explanation

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Ionic and Molecular Compounds

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