# 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?

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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**