If 1.00 mol of argon is placed in a 0.500-l container at 20.0 ∘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.

Hey There!

ideal gas law  :

PV = nRT

P = nRT/V

P = (1*0.082)(18+273)/(0.5) = 47.724 atm

For VDW :

(P + a(n/V)²)(V - nB) = nRT

P = nRT/(V - nB) - a(n/V)²

P = 1*0.082*(18+273)  / (0.5-1*0.03219) - 1.345*( 1/0.5 )² = 45.62

P = 45.62 atm

Pdif = P2-P1 = 47.724 - 45.62 = 2.104 atm

if the concentration of products is decreased, the equilibrium will shift toward the products). add an inert gas (one that is not involved in the reaction) to the constant-volume reaction mixture: this will increase the total pressure of the system, but will have no effect on the equilibrium condition.

explanation: