Use the van der Waals equation of state to calculate the pressure of 3.70 mol of SO, at 467K in a 3.60 L vessel. Van der Waals constants can be found in the van der Waals constants table. P = atm Use the ideal gas equation to calculate the pressure under the same conditions. P = atm In a 15.40 L vessel, the pressure of 3.70 mol of SO, at 467 K is 9.21 atm when calculated using the ideal gas equation and 8.94 atm when calculated using the van der Waals equation of state. Why is the percent difference in the pressures calculated using the two different equations greater when the gas is in the 3.60 L vessel compared to the 15.40 L vessel? The attractive forces between molecules become a greater factor at the higher pressure in the 3.60L vessel. The molecular volume is a larger part of the total volume of the 3.60L vessel. The attractive forces between molecules become less of a factor at the higher pressure in the 3.60L vessel. The molecular volume is a smaller part of the total volume of the 3.6OL vessel.

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...
icon
Related questions
Question
Use the van der Waals equation of state to calculate the pressure of 3.70 mol of SO, at 467K in a 3.60 L vessel. Van der Waals
constants can be found in the van der Waals constants table.
P =
atm
Use the ideal gas equation to calculate the pressure under the same conditions.
P =
atm
In a 15.40 L vessel, the pressure of 3.70 mol of SO, at 467 K is 9.21 atm when calculated using the ideal gas equation and
8.94 atm when calculated using the van der Waals equation of state.
Why is the percent difference in the pressures calculated using the two different equations greater when the gas is in the
3.60 L vessel compared to the 15.40 L vessel?
O The attractive forces between molecules become a greater factor at the higher pressure in the 3.60 L vessel.
The molecular volume is a larger part of the total volume of the 3.60L vessel.
The attractive forces between molecules become less of a factor at the higher pressure in the 3.60L vessel.
The molecular volume is a smaller part of the total volume of the 3.6OL vessel.
étv
Transcribed Image Text:Use the van der Waals equation of state to calculate the pressure of 3.70 mol of SO, at 467K in a 3.60 L vessel. Van der Waals constants can be found in the van der Waals constants table. P = atm Use the ideal gas equation to calculate the pressure under the same conditions. P = atm In a 15.40 L vessel, the pressure of 3.70 mol of SO, at 467 K is 9.21 atm when calculated using the ideal gas equation and 8.94 atm when calculated using the van der Waals equation of state. Why is the percent difference in the pressures calculated using the two different equations greater when the gas is in the 3.60 L vessel compared to the 15.40 L vessel? O The attractive forces between molecules become a greater factor at the higher pressure in the 3.60 L vessel. The molecular volume is a larger part of the total volume of the 3.60L vessel. The attractive forces between molecules become less of a factor at the higher pressure in the 3.60L vessel. The molecular volume is a smaller part of the total volume of the 3.6OL vessel. étv
Van der Waals Constants
la
Gas
la
b (L/mol)
(bar L2 /mol 2) (atm L2 /mol 2
2.303
CH
2.273
0.0431
CO2
Cl2
NH3
H20
Xe
3.658
3.610
0.0429
6.343
6.260
0.0542
4.225
4.170
0.0371
5.537
4.192
5.465
0.0305
4.137
0.0516
20.01
19.75
0.1281
02
N2
1.382
1.363
0.0319
1.370
1.351
0.0387
Kr
Ar
5.193
5.121
0.0106
1.355
1.336
0.0320
CO
1.472
1.452
0.0395
H2S
4.544
4.481
0.0434
NO
1.370
3.852
5.36
1.351
0.0387
N20
NO2
SO2
3.799
0.0444
5.29
0.0443
6.865
6.770
9.431
3.648
4.437
6,221
0.0568
HF
9.565
3.700
0,0739
HCI
HBr
HI
0.0406
0.0442
4.500
6.309
0.0530
Transcribed Image Text:Van der Waals Constants la Gas la b (L/mol) (bar L2 /mol 2) (atm L2 /mol 2 2.303 CH 2.273 0.0431 CO2 Cl2 NH3 H20 Xe 3.658 3.610 0.0429 6.343 6.260 0.0542 4.225 4.170 0.0371 5.537 4.192 5.465 0.0305 4.137 0.0516 20.01 19.75 0.1281 02 N2 1.382 1.363 0.0319 1.370 1.351 0.0387 Kr Ar 5.193 5.121 0.0106 1.355 1.336 0.0320 CO 1.472 1.452 0.0395 H2S 4.544 4.481 0.0434 NO 1.370 3.852 5.36 1.351 0.0387 N20 NO2 SO2 3.799 0.0444 5.29 0.0443 6.865 6.770 9.431 3.648 4.437 6,221 0.0568 HF 9.565 3.700 0,0739 HCI HBr HI 0.0406 0.0442 4.500 6.309 0.0530
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 3 steps with 2 images

Blurred answer
Knowledge Booster
Absorption and Adsorption
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
Chemistry
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Organic Chemistry
Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning
Elementary Principles of Chemical Processes, Bind…
Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY