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.

d is

a). in this case, dh> 0 and sd> 0. if t becomes larger, the term "-tds" will become more negative. at some point, it is possible that -dts will be more negative than the dh is positive, and the reaction will become spontaneous. this is probably the correct answer, but you should look at all of the other possibilities and make sure of that.