Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 60

It takes \(12.6 \mathrm{~s}\) for \(1.73 \times 10^{-3} \mathrm{~mol}\) of \(\mathrm{CO}\) to effuse through a pinhole. Under the same conditions, how long will it take for the same amount of \(\mathrm{CO}_{2}\) to effuse through the same pinhole?

Short Answer

Expert verified
Answer: It will take approximately 10.08 seconds for the same amount of CO2 to effuse through the same pinhole.

Step by step solution

01

Identify necessary information

First, we should gather the needed information to solve the problem. From the exercise, we know the following: - It takes 12.6 seconds for \(1.73 \times 10^{-3} \mathrm{~mol}\) of CO to effuse through a pinhole. - We need to determine the time it takes for the same amount (\(1.73 \times 10^{-3} \mathrm{~mol}\)) of CO2 to effuse through the same pinhole. - We'll use Graham's Law of Effusion to relate the effusion rates of CO and CO2.
02

Utilize Graham's Law of Effusion

Graham's Law of Effusion states that the ratio of the effusion rates of two gases is inversely proportional to the square root of their molar masses. Mathematically, this can be represented as: \(\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}\) where \(r_1\) and \(r_2\) are the effusion rates of gas 1 and gas 2, and \(M_1\) and \(M_2\) are their molar masses. In our case, gas 1 is CO and gas 2 is CO2.
03

Calculate molar masses of CO and CO2

Next, we need to find the molar masses of CO and CO2. For CO: Molar mass of C = 12.01 g/mol Molar mass of O = 16.00 g/mol Molar mass of CO = \((12.01 + 16.00) \mathrm{~g/mol} = 28.01 \mathrm{~g/mol}\) For CO2: Molar mass of C = 12.01 g/mol Molar mass of O = 16.00 g/mol Molar mass of CO2 = \((12.01 + 2(16.00)) \mathrm{~g/mol} = 44.01 \mathrm{~g/mol}\)
04

Apply Graham's Law formula to compute effusion rate ratio

Now we can apply the formula of Graham's Law for effusion rate ratio. By plugging in the molar masses for CO and CO2, we get: \(\frac{r_{CO}}{r_{CO2}} = \sqrt{\frac{M_{CO2}}{M_{CO}}} = \sqrt{\frac{44.01}{28.01}}\) Now, compute the ratio: \(\frac{r_{CO}}{r_{CO2}} \approx 1.25\)
05

Find the time it takes for CO2 to effuse through the pinhole

Recall that the effusion rate is equal to the number of moles divided by the time (r = moles/time). From the exercise, we know it takes 12.6 seconds for \(1.73 \times 10^{-3} \mathrm{~mol}\) of CO to effuse. Since effusion rates are inversely proportional to time, we can rearrange our equation and relate the amount of time for CO and CO2 to effuse as follows: \(\frac{t_{CO}}{t_{CO2}} \approx 1.25\) By plugging in the given time for CO, we can find the time for CO2 effusing through the pinhole: \(t_{CO2} = \frac{t_{CO}}{1.25} \approx \frac{12.6 \mathrm{~s}}{1.25} \approx 10.08 \mathrm{~s}\)
06

Final Answer

Under the same conditions, it will take approximately \(10.08 \mathrm{~s}\) for the same amount of CO2 to effuse through the same pinhole.

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

Hydrogen sulfide gas \(\left(\mathrm{H}_{2} \mathrm{~S}\right)\) is responsible for the foul odor of rotten eggs. When it reacts with oxygen, sulfur dioxide gas and steam are produced. (a) Write a balanced equation for the reaction. (b) How many liters of \(\mathrm{H}_{2} \mathrm{~S}\) would be required to react with excess oxygen to produce \(12.0 \mathrm{~L}\) of \(\mathrm{SO}_{2}\) ? The reaction yield is \(88.5 \%\). Assume constant temperature and pressure throughout the reaction.

Glycine is an amino acid made up of carbon, hydrogen, oxygen, and nitrogen atoms. Combustion of a \(0.2036\) -g sample gives \(132.9 \mathrm{~mL}\) of \(\mathrm{CO}_{2}\) at \(25^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) and \(0.122 \mathrm{~g}\) of water. What are the percentages of carbon and hydrogen in glycine? Another sample of glycine weighing \(0.2500 \mathrm{~g}\) is treated in such a way that all the nitrogen atoms are converted to \(\mathrm{N}_{2}(g)\). This gas has a volume of \(40.8 \mathrm{~mL}\) at \(25^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\). What is the percentage of nitrogen in glycine? What is the percentage of oxygen? What is the empirical formula of glycine?

A sample of \(\mathrm{CO}_{2}\) gas at \(22^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) has a volume of \(2.00 \mathrm{~L}\). Determine the ratio of the original volume to the final volume when (a) the pressure and amount of gas remain unchanged and the Celsius temperature is doubled. (b) the pressure and amount of gas remain unchanged and the Kelvin temperature is doubled.

Some chambers used to grow bacteria that thrive on \(\mathrm{CO}_{2}\) have a gas mixture consisting of \(95.0 \% \mathrm{CO}_{2}\) and \(5.0 \% \mathrm{O}_{2}\) (mole percent). What is the partial pressure of each gas if the total pressure is \(735 \mathrm{~mm} \mathrm{Hg}\) ?

A 1.58-g sample of \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{X}_{3}(g)\) has a volume of \(297 \mathrm{~mL}\) at \(769 \mathrm{~mm} \mathrm{Hg}\) and \(35^{\circ} \mathrm{C}\). Identify the element \(\mathrm{X}\).
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

The Earths Atmosphere

Read Explanation

Physical Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Kinetics

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