The observed pressure of a 1.385-mole sample of Kr(g) in a 9.338-L container is 2.826 atm....
The observed pressure of a 2.963-mole sample of Kr(g) in a 9.450-L container is 1.782 atm. Use the van der Waals equation to determine the pressure if the gas were behaving ideally. Kr(s) = 2.32 atm. L2 mol Our(s) = Ideal pressure atm
Hint: % difference = 100×(P ideal - Pvan der Waals) / P idealAccording to the ideal gas law, a 9.843 mol sample of argon gas in a 0.8425 L container at 502.0 K should exert a pressure of 481.3 atm. By what percent does the pressure calculated using the van der Waals' equation differ from the ideal pressure? For Ar gas, a =1.345L2 atm/mol2 and b = 3.219×10-2 L/mol.
A 1.55-mol sample of nitrogen gas is maintained in a 0.730-L container at 292 K. Calculate the pressure of the gas using both the ideal gas law and the van der Waals equation (van der Waals constants for N2 are a = 1.39 L2atm/mol2 and b = 3.91×10-2 L/mol). Pideal gas equation = ______ atm Pvan der Waals =_____ atm
A 9.450 mol sample of krypton gas is maintained in a 0.8100 L container at 300.1 K. What is the pressure in atm calculated using the van der Waals' equation for Kr gas under these conditions? For Kr, a = 2.318 L2atm/mol2 and b = 3.978×10-2 L/mol. atm
A 10.13 mol sample of krypton gas is maintained in a 0.7517 L container at 297.2 K. What is the pressure in atm calculated using the van der Waals' equation for Kr gas under these conditions? For Kr, a = 2.318 L2atm/mol2 and b = 3.978×10-2 L/mol.
If 1.00 mol of argon is placed in a 0.500-L container at 30.0 ∘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.
According to the ideal gas law, a 1.066 mol sample of krypton gas in a 1.927 L container at 272.4 K should exert a pressure of 12.37 atm. By what percent does the pressure calculated using the van der Waals' equation differ from the ideal pressure? For Kr gas, a = 2.318 L2atm/mol2 and b = 3.978×10-2 L/mol. ___% Hint: % difference = 100×(P ideal - Pvan der Waals) / P ideal
A 10.83 mol sample of argon gas is maintained in a 0.7598 L container at 295.6 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 Llatm/mol2 and b = 3.219x10-2 L/mol. atm USE LIIC CICICULUS LU al USS IITIPUI lalil va The rate of effusion of H2 gas through a porous barrier is observed to be 1.17 x 10-4 mol/h. Under the same conditions,...
According to the ideal gas law, a 10.74 mol sample of krypton gas in a 0.8444 L container at 498.7 K should exert a pressure of 520.5 atm. What is the percent difference between the pressure calculated using the van der Waals' equation and the ideal pressure? For Kr gas, a = 2.318 L2atm/mol2 and b = 3.978×10-2 L/mol. ____%
If 1.00 mol of argon is placed in a 0.500-L container at 30.0 ∘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.