Suppose 10.00 mol of Ar gas are pumped into a 1.01 L container at 293.15 K.
Calculate the expected pressure based on the ideal gas law, Pideal.
?atm
Calculate an estimate of the gas pressure one might observe
based on the van der Waals equation,
Pobs.
a = 1.34 L2 · atm/mol2, b =
0.0322 L/mol.
atm
Part 3 (1 point)
Which factor, a or b, is the primary reason
for the difference in the calculated pressures?
Choose one:
a.
b.
What does the factor selected in Part 3 reflect?
Choose one:
A. intermolecular attractions
B. volume occupied by the molecules
C. intermolecular repulsions
D. low average kinetic energy
Suppose 10.00 mol of Ar gas are pumped into a 1.01 L container at 293.15 K....
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
The van der Waals equation gives a relationship between the pressure p (atm), volume V(L), and temperature T(K) for a real gas: .2 where n is the number of moles, R 0.08206(L atm)(mol K) is the gas con- stant, and a (L- atm/mol-) and b (L/mol) are material constants. Determine the volume of 1.5 mol of nitrogen (a .39 L2 atm/mol2. b = 0.03913 L/mol) at temperature of 350 K and pressure of 70 atm. The van der Waals equation...
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.
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 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.
A 9.262 mol sample of xenon gas is maintained in a 0.8496 L container at 299.3 K. What is the pressure in atm calculated using the van der Waals' equation for Xe gas under these conditions? For Xe, a = 4.194 L2atm/mol2 and b = 5.105×10-2 L/mol.
A 9.386 mol sample of methane gas is maintained in a 0.7854 L container at 302.4 K. What is the pressure in atm calculated using the van der Waals' equation for CH4 gas under these conditions? For CH4, a = 2.253 L2atm/mol2 and b = 4.278×10-2 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.
If 1.00 mol of argon is placed in a 0.500-L container at 22.0 ∘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.