Homework Help is Here – Start Your Trial Now!
Science
Chemistry
Example #3: 1.85 L of a gas is collected over water at 98.0 kPa and 22.0 °C. What is the volume of the dry gas at STP?

Example #3: 1.85 L of a gas is collected over water at 98.0 kPa and 22.0 °C. What is the volume of the dry gas at STP?

General Chemistry - Standalone book (MindTap Course List)
General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Chapter5: The Gaseous State
Section: Chapter Questions
Problem 5.127QP: A 1.000-g sample of an unknown gas at 0C gives the following data: P(atm) V (L) 0.2500 3.1908 0.5000...
See similar textbooks
icon
Related questions
Question
Explain the steps used to solve Example #3:
The next example uses two gas laws in sequence. It involves using Dalton's Law of Partial Pressures first, then use of the Combined Gas Law. The explanation will assume you understand Dalton's Law. These two laws occuring together in a problem is VERY COMMON. Example #3: 1.85 L of a gas is collected over water at 98.0 kPa and 22.0 °C. What is the volume of the dry gas at STP? The key phrase is "over water." Another common phrase used in this type of problem is "wet gas." This means the gas was collected by bubbling it into an inverted bottle filled with water which is sitting in a water bath. The gas bubbles in and is trapped. It displaces the water which flows out into the water bath. The terms "over water" and "wet gas" are equivalent; they means the same thing, that being that the gas is saturated with water vapor. The problem is that the trapped gas now has water vapor mixed in with it. This is a consequence of the technique and cannot be avoided. However, there is a calculation technique (Dalton's Law) that allows use to remove the effect of the water vapor and treat the gas as "dry." For this example, we write Dalton's Law like this: Pgas + PH,0 = Ptotal We need to know the vapor pressure of water at 22.0 °C and to do this we must look it up in a reference source. It is important to recognize the Ptotal is the 98.0 value. Ptotal is the combined pressure of the dry gas AND the water vapor. We want the water vapor's pressure OUT. We put the values into the Dalton's Law equation: Pgas + 2.6447 kPa = 98.0 kPa We solve the problem for Pgas and get 95.3553 kPa. Notice that it is not rounded off. The only rounding off done is at the FINAL answer, which this is not. Placing all the values into the solution matrix yields this: P =95.3553 kPa P, = 101.325 kPa V1 = 1.85 L V2 x T = 295 K T2 = 273 K Solve for x in the usual manner of cross-multiplying and dividing: V2 (PIV1T2) / (P2T1) x = [(95.3553 kPa) (1.85 L) (273 K)] / [(101.325 kPa) (295 K) x = 1.61 L (to three sig figs) Comment: a very common student mistake is to not realize that Dalton's Law must be used first when a gas is collected over water. There is a very common experiment in which some hydrogen gas is collected over water and the molar volume is determined. Dalton's Law will be used in the calculations associated with that lab.
Transcribed Image Text:The next example uses two gas laws in sequence. It involves using Dalton's Law of Partial Pressures first, then use of the Combined Gas Law. The explanation will assume you understand Dalton's Law. These two laws occuring together in a problem is VERY COMMON. Example #3: 1.85 L of a gas is collected over water at 98.0 kPa and 22.0 °C. What is the volume of the dry gas at STP? The key phrase is "over water." Another common phrase used in this type of problem is "wet gas." This means the gas was collected by bubbling it into an inverted bottle filled with water which is sitting in a water bath. The gas bubbles in and is trapped. It displaces the water which flows out into the water bath. The terms "over water" and "wet gas" are equivalent; they means the same thing, that being that the gas is saturated with water vapor. The problem is that the trapped gas now has water vapor mixed in with it. This is a consequence of the technique and cannot be avoided. However, there is a calculation technique (Dalton's Law) that allows use to remove the effect of the water vapor and treat the gas as "dry." For this example, we write Dalton's Law like this: Pgas + PH,0 = Ptotal We need to know the vapor pressure of water at 22.0 °C and to do this we must look it up in a reference source. It is important to recognize the Ptotal is the 98.0 value. Ptotal is the combined pressure of the dry gas AND the water vapor. We want the water vapor's pressure OUT. We put the values into the Dalton's Law equation: Pgas + 2.6447 kPa = 98.0 kPa We solve the problem for Pgas and get 95.3553 kPa. Notice that it is not rounded off. The only rounding off done is at the FINAL answer, which this is not. Placing all the values into the solution matrix yields this: P =95.3553 kPa P, = 101.325 kPa V1 = 1.85 L V2 x T = 295 K T2 = 273 K Solve for x in the usual manner of cross-multiplying and dividing: V2 (PIV1T2) / (P2T1) x = [(95.3553 kPa) (1.85 L) (273 K)] / [(101.325 kPa) (295 K) x = 1.61 L (to three sig figs) Comment: a very common student mistake is to not realize that Dalton's Law must be used first when a gas is collected over water. There is a very common experiment in which some hydrogen gas is collected over water and the molar volume is determined. Dalton's Law will be used in the calculations associated with that lab.
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Mole Concept
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
Recommended textbooks for you
General Chemistry - Standalone book (MindTap Cour…
General Chemistry - Standalone book (MindTap Cour…
Chemistry
ISBN:
9781305580343
Author:
Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:
Cengage Learning
Physical Chemistry
Physical Chemistry
Chemistry
ISBN:
9781133958437
Author:
Ball, David W. (david Warren), BAER, Tomas
Publisher:
Wadsworth Cengage Learning,
Chemistry
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Introduction to General, Organic and Biochemistry
Introduction to General, Organic and Biochemistry
Chemistry
ISBN:
9781285869759
Author:
Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Publisher:
Cengage Learning
Chemistry: An Atoms First Approach
Chemistry: An Atoms First Approach
Chemistry
ISBN:
9781305079243
Author:
Steven S. Zumdahl, Susan A. Zumdahl
Publisher:
Cengage Learning
Chemistry for Engineering Students
Chemistry for Engineering Students
Chemistry
ISBN:
9781337398909
Author:
Lawrence S. Brown, Tom Holme
Publisher:
Cengage Learning
SEE MORE TEXTBOOKS