If 1.00 mol of argon is placed in a 0.500-L container at 19.0 ∘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.

41.083 atm 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.

Explanation:

Data given for argon gas:

number of moles = 1 mole

volume = 0.5 L

Temperature = 19 degrees or 292.15 K

a= 1.345 (L2⋅atm)/mol2

b= 0.03219L/mol.

R = 0.0821

The real pressure equation given by Van der Waals equation:

P =( RT ÷ Vm-b) - a ÷ Vm^2

Putting the values in the equation:

P = (0.0821 x 292.15) ÷(0.5 - 0.03219) - 1.345÷ (0.5)^2

  = 23.98÷0.4678 - 1.345 ÷0 .25

  = 51.26 - 5.38

  = 45.88 atm is the real pressure.

The pressure from the ideal gas law

PV =nRT

P =( 1 x 0.0821 x 292.15) ÷ 0.5

  = 4.797 atm

the difference between the ideal pressure and real pressure is

Pressure by vander waal equation- Pressure by ideal gas law

45.88 - 4.797

= 41.083 atm.is the difference between the two.

sry but any answer choices

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