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If 1.00 mol of argon is placed in a 0.500- L container at 24.0 ∘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.

Sagot :

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The difference between the ideal pressure and the pressure calculated by the Van Der Waal equation is 2.08 atm.

What is the pressure?

In this problem, we are mandated to obtain the pressure both by the use of the ideal gas equation and then the use of the Van der Walls equation.

Using the idea gas equation;

PV = nRT

P = nRT/V

P = pressure

V = volume

n = number of moles

T = temperature

R = gas constant

P = 1 * 0.082 * (24 + 273)/0.5

P = 48.7 atm

Using the Van Der Wall equation:

P = RT/(V - b) - a /V^2

P = 0.082 * 297/(0.5 - 0.03219) - 1.345/(0.5)^2

P = 24.354/0.46781 - 1.345/ /0.25

P = 52 - 5.38

P = 46.62 atm

The difference between the ideal pressure and the pressure calculated by the Van Der Waal equation is; 48.7 atm - 46.62 atm = 2.08 atm

Learn more about ideal gas equation:https://brainly.com/question/3637553

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