According to the ideal gas law a 9.03 sample of oxygen gas in a 0.8112 container at 499k should exert a pressure of 456atm. By what % does the pressure calculated using van der waals equation differ from the ideal pressure?

For O2 gas a=1.36 L^2atm/mol2 and b=3.18e-2

Hint: %diff= 100 X (P ideal - P van der waal) / P ideal

I know that van der waals equation is
(P + a [n/v]^2) (v-bn) = nRT but I have no idea how to use this and my textbook does not give any examples. Thank you for your help!

First, you omitted the units on the 9.03 WHAT in a 0.8112 WHAT container. I'm confused because you are confused. There is nothing here that you need an idea about. You have a and b, R, T, and n. Solve for P by van d equation, then use the usual PV = nRT and solve for P. Then follow the hint in the problem which tells you how to solve for the % difference.

To find the percentage difference between the pressure calculated using the ideal gas law and the Van der Waals equation, you need to follow these steps:

Step 1: Calculate the pressure using the ideal gas law.
You can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

Given:
V = 0.8112 L (volume of the container)
n = 9.03 g / (32.00 g/mol) = 0.28219 mol (number of moles)
T = 499 K (temperature)
R = 0.0821 L·atm/mol·K (gas constant for atm unit)

Plugging the values into the ideal gas law equation:
P_ideal = nRT / V
P_ideal = (0.28219 mol)(0.0821 L·atm/mol·K)(499 K) / (0.8112 L)
P_ideal = 11.9078 atm

So, the pressure calculated using the ideal gas law is 11.9078 atm.

Step 2: Calculate the pressure using the Van der Waals equation.
The Van der Waals equation accounts for the volume of the gas molecules (b) and the intermolecular forces (a) that affect the behavior of real gases.

Given:
a = 1.36 L^2·atm/mol^2
b = 3.18 x 10^-2 L/mol

Plugging the values into the Van der Waals equation:
(P + a[n/V]^2)(V - bn) = nRT
PVan_der_Waals = (P + (a[n/V]^2))(V - bn) / V
Note: The term n/V is the molar concentration, which can be calculated as n/V = P/(RT).

Calculating n/V:
(n/V) = P / (RT)
(n/V) = 456 atm / ((0.0821 L·atm/mol·K)(499 K))
(n/V) = 0.10968 mol/L

Plugging the values into the Van der Waals equation:
PVan_der_Waals = (456 atm + (1.36 L^2·atm/mol^2)((0.10968 mol/L) / (0.8112 L))^2)(0.8112 L - (3.18 x 10^-2 L/mol)(0.10968 mol/L)) / 0.8112 L
PVan_der_Waals = 13.488 atm

So, the pressure calculated using the Van der Waals equation is 13.488 atm.

Step 3: Calculate the percentage difference.
The percentage difference between the two pressures can be found using the given formula:
% diff = 100 * (P_ideal - PVan_der_Waals) / P_ideal

Plugging in the values:
% diff = 100 * (11.9078 atm - 13.488 atm) / 11.9078 atm
% diff = -13.303%

Therefore, the pressure calculated using the Van der Waals equation differs from the ideal pressure by approximately 13.303%.