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?
For argon, a=1.345(L2⋅atm)/mol2 and b=0.03219L/mol.
For argon, a=1.345(L2⋅atm)/mol2 and b=0.03219L/mol.
1 Answer
Jan 22, 2018
2.03 atm
Explanation:
Firstly, lets calculate the ideal gas equation :
Now, plug in numbers
Real gas equation
Plug in numbers
P (real) = 46.57 atm
P(ideal) - P(real) = 2.03 atm