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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?

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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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.
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