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13 Dec 2019

If 1.00 mol of argon is placed in a 0.500-L container at 20.0 âC , what is the difference between the ideal pressure (as predicted by the ideal gas law) and the real pressure (as predicted by the van der Waals equation)? For argon, a=1.345(L2â atm)/mol2 and b=0.03219L/mol. Enter the pressure difference in atmospheres using two significant figures.

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17 Dec 2019

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If 1.00 mol of argon is placed in a 0.500-L container at 28.0 ?C , what is the difference between the ideal pressure (as predicted by the ideal gas law) and the real pressure (as predicted by the van der Waals equation)? For argon, a=1.363L2 bar mol?2 and b=0.03219L mol?1.

Part A

A 3.00-L flask is filled with gaseous ammonia, NH3. The gas pressure measured at 26.0?C is 2.05atm . Assuming ideal gas behavior, how many grams of ammonia are in the flask?

Express your answer to three significant figures and include the appropriate units.

mass of NH3 =

Part B

If 1.00 mol of argon is placed in a 0.500-L container at 29.0?C , what is the difference between the ideal pressure (as predicted by the ideal gas law) and the real pressure (as predicted by the van der Waals equation)?

For argon, a=1.345(L2?atm)/mol2 and b=0.03219L/mol.

Express your answer to two significant figures and include the appropriate units.

Pideal?Preal =

1) A 10.60 mol sample of argon gas is maintained in a 0.8466 L container at 298.6 K. What is the pressure in atm calculated using the van der Waals' equation for Ar gas under these conditions? For Ar, a = 1.345 L²atm/mol² and b = 3.219Ã10^-2 L/mol.

A: ______ atm

2) According to the ideal gas law, a 1.007 mol sample of argon gas in a 1.337 L container at 272.3 K should exert a pressure of 16.83 atm. By what percent does the pressure calculated using the van der Waals' equation differ from the ideal pressure? For Ar gas, a = 1.345 L²atm/mol² and b = 3.219Ã10^-2L/mol.

Hint: % difference = 100Ã(P _ideal - P_{van der Waals}) / P _ideal

A: ________% â

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