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Use the van der Waals equation of state to calculate the pressure of 2.00 mol of CH, at 469 K in a 5.90 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

Use the van der Waals equation of state to calculate the pressure of 2.00 mol of CH, at 469 K in a 5.90 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

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...
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Use the van der Waals equation of state to calculate the pressure of 2.00 mol of CH, at 469 K in a 5.90 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 Under these conditions, would you expect CH, or CCI, to deviate more from ideal behavior? Why? CCI because it occupies a smaller volume and it has greater dispersion forces between molecules. O CH, because it occupies a smaller volume and it has greater dispersion forces between molecules. CH because it occupies a smaller volume and it has smaller dispersion forces between molecules. CCI because it occupies a larger volume and it has smaller dispersion forces between molecules. O CH because it occupies a larger volume and it has smaller dispersion forces þetween molecules. O CCI, because it occupies a larger volume and it has greater dispersion forces between molecules. O O O C O0
Transcribed Image Text:Use the van der Waals equation of state to calculate the pressure of 2.00 mol of CH, at 469 K in a 5.90 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 Under these conditions, would you expect CH, or CCI, to deviate more from ideal behavior? Why? CCI because it occupies a smaller volume and it has greater dispersion forces between molecules. O CH, because it occupies a smaller volume and it has greater dispersion forces between molecules. CH because it occupies a smaller volume and it has smaller dispersion forces between molecules. CCI because it occupies a larger volume and it has smaller dispersion forces between molecules. O CH because it occupies a larger volume and it has smaller dispersion forces þetween molecules. O CCI, because it occupies a larger volume and it has greater dispersion forces between molecules. O O O C O0
Van der Waals Constants la Gas b (L/mol) (bar L2 /mol 2 ) (atm L2 /mol 2) 2.303 CHA CO2 Cl2 NH3 H20 2.273 0.0431 3.658 3.610 0.0429 6.343 6.260 0.0542 4.225 4.170 0.0371 5.537 5.465 0,0305 Xe 4.192 4.137 19.75 0.0516 CCI4 02 N2 20.01 0.1281 1.382 1.363 0.0319 1.370 1.351 0.0387 Kr 5.193 5.121 1.336 1.452 0.0106 Ar 1.355 1.472 0.0320 CO 0.0395 H,S NO N20 NO2 SO2 4.544 1.370 4.481 0.0434 1.351 0.0387 3.852 3.799 0.0444 5.36 5.29 0.0443 6.865 6.770 0.0568 HF 9.565 3.700 4.500 6.309 9.431 0.0739 HCI HBr 3.648 4.437 6.221 0.0406 0.0442 0.0530 HI
Transcribed Image Text:Van der Waals Constants la Gas b (L/mol) (bar L2 /mol 2 ) (atm L2 /mol 2) 2.303 CHA CO2 Cl2 NH3 H20 2.273 0.0431 3.658 3.610 0.0429 6.343 6.260 0.0542 4.225 4.170 0.0371 5.537 5.465 0,0305 Xe 4.192 4.137 19.75 0.0516 CCI4 02 N2 20.01 0.1281 1.382 1.363 0.0319 1.370 1.351 0.0387 Kr 5.193 5.121 1.336 1.452 0.0106 Ar 1.355 1.472 0.0320 CO 0.0395 H,S NO N20 NO2 SO2 4.544 1.370 4.481 0.0434 1.351 0.0387 3.852 3.799 0.0444 5.36 5.29 0.0443 6.865 6.770 0.0568 HF 9.565 3.700 4.500 6.309 9.431 0.0739 HCI HBr 3.648 4.437 6.221 0.0406 0.0442 0.0530 HI
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