If 1.00 mol of argon is placed in a 0.500-L container at 23.0 ∘C , what is the difference between the ideal pressure and the real pressure?

Question

For argon, a=1.345(L2⋅atm)/mol2 and b=0.03219L/mol.

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Answer 1 · 2018-01-22T02:29:47

2.03 atm

Firstly, lets calculate the ideal gas equation :

#P=(n*R*T)/V#

Now, plug in numbers

#P=(1*0.0821*296K)/(0.5L)=48.6 atm#

Real gas equation

#(P+(a*n^2)/V^2) * (V-nb) = nRT#

Plug in numbers

#(P+(1.345*1^2)/0.5^2) * (0.5-1*0.03219) = 1*0.0821*296#

P (real) = 46.57 atm

P(ideal) - P(real) = 2.03 atm