If 1.00mol of argon is placed in a 0.500- container at 28.0 degree Celsius, 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(L square. atm)/mol square and b=0.03219L/mol .

You know the ideal PV = nRT equation, and you know van der Waals equation. Plug and chug.

To find the difference between the ideal pressure and the real pressure, we need to use the ideal gas law and the van der Waals equation.

1. The ideal gas law is given by the equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin.

2. For a given gas, the van der Waals equation is given by: (P + a(n/V)^2)(V - nb) = nRT, where a and b are constants specific to the gas.

Given:
n = 1.00 mol (number of moles of argon)
V = 0.500 L (volume of the container)
T = 28.0 + 273.15 = 301.15 K (temperature in Kelvin)
a = 1.345 (L^2⋅atm)/(mol^2) (constant for argon)
b = 0.03219 L/mol (constant for argon)

3. Calculating the ideal pressure (P_ideal) using the ideal gas law:
Using PV = nRT, we can solve for P:
P_ideal = (nRT) / V

Plug in the given values:
P_ideal = (1.00 mol * 0.0821 L·atm/mol·K * 301.15 K) / 0.500 L
P_ideal ≈ 49.64 atm

4. Calculating the real pressure (P_real) using the van der Waals equation:
Using (P + a(n/V)^2)(V - nb) = nRT, we can solve for P:
P_real = (nRT) / (V - nb) - a(n/V)^2

Plug in the given values:
P_real = (1.00 mol * 0.0821 L·atm/mol·K * 301.15 K) / (0.500 L - 1.00 mol * 0.03219 L/mol)
P_real ≈ 48.58 atm

5. Finding the difference between the ideal pressure and the real pressure:
Difference = P_ideal - P_real
Difference ≈ 49.64 atm - 48.58 atm
Difference ≈ 1.06 atm

Therefore, the difference between the ideal pressure and the real pressure is approximately 1.06 atm.