Question

#25

A 10.86 mol sample of krypton gas is maintained in a 0.7529 L container at 296.3 K. What is the pressure in atm calculated using the van der Waals' equation for Kr gas under these conditions? For Kr, a = 2.318L²atm/mol² and b = 3.978×10^-2 L/mol.

____atm

According to the ideal gas law, a 1.077 mol sample of methane gas in a 1.670 L container at 265.4 K should exert a pressure of 14.05 atm. By what percent does the pressure calculated using the van der Waals' equation differ from the ideal pressure? For CH₄ gas, a = 2.253 L²atm/mol² and b = 4.278×10^-2 L/mol.

______%

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

According to the ideal gas law, a 9.308 mol sample of nitrogen gas in a 0.8235 L container at 502.7 K should exert a pressure of 466.3 atm. By what percent does the pressure calculated using the van der Waals' equation differ from the ideal pressure? For N₂ gas, a = 1.390 L²atm/mol² and b = 3.910×10^-2 L/mol.

%

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

.26

A 9.65×10^-2 mol sample of an unknown gas contained in a 5.00 L flask is found to have a density of 0.720 g/L. The molecular weight of the unknown gas is _____g/mol.

A 7.01×10^-2 mol sample of Ne gas is contained in a 2.00 L flask at room temperature and pressure. What is the density of the gas, in grams/liter, under these conditions?

_______g/L

What is the molar volume of Xe gas under the conditions of temperature and pressure where its density is 2.84 g/L?

_____ L/mol

Answer 1

25:

krypton gas :

We know the van der Waals' gas equation = [P + a(n/V)^2] [V/n-b] = RT

From the given data a = 2.318 L²atm/mol² and b = 3.978×10^-2 L/mol

Volume V= 0.7529 L, T = 296.3 K, n = 10.86 moles

if plugin all the above values in the equation

[P+ 2.318 L²atm/mol² (10.86 moles/0.7529 L)^2] [0.7529 L/10.86 moles - 3.978×10^-2 L/mol] = 0.0821 L.atm /mole.K x 296.3 K

[P + 482.2791 atm] [2.9547 x 10-2 L/mol] = 24.32623 L.atm/mole

[P + 482.2791 atm] = 8.23306 x 10² atm

P = 341.0269 atm

Methane gas :

Similarly if we calculated for CH4 the pressure calculated using the van der Waals' equation

[P+ 2.253 L²atm/mol² (1.077 moles/1.670 L)^2] [1.670 L/1.077 moles - 4.278×10^-2 L/mol] = 0.0821 L.atm /mole.K x 265.4 K

P = 13.514 atm

Ideal pressure Pi = 14.05 atm

% difference = 100×(P _ideal - P_{van der Waals}) / P _ideal = 100 (14.05-13.514)/14.05 = 3.815 %

Nitrogen gas :

[P+ 1.390 L²atm/mol² (9.308 moles/0.8235 L)^2] [0.8235 L/9.308 moles - 3.910×10^-2 L/mol] = 0.0821 L.atm /mole.K x 502.7 K

177.5825 x 0.04937 =

P = 658.3840 atm

Ideal pressure Pi = 466.3 atm

% difference = 100×(P _ideal - P_{van der Waals}) / P _ideal = 100 (14.05-13.514)/14.05 = - 41.19 %