Question

If 1.00 mol of argon is placed in a 0.500-L container at 29.0 ∘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.

Answer 1

Ideal ,P=49.52 atm

Real ,P=47.62 atm

Step-by-step explanation:

Given that

n= 1 mol

V= 0.5 L

T= 29 ∘C = 29 +273 K

T= 302 K

For ideal gas

P V = n R T

P x 0.5 = 1 x 0.0821 x 302

P=49.52 atm

For real gas

`\left ( P+(an^2)/(v^2) \right )\left ( v-nb \right )=nRT`

Now by putting the values

`\left ( P+(1.345* 1^2)/(0.5^2) \right )\left ( 0.5-1* 0.03219 \right )=1* 0.0821* 302`

`\left ( P+(1.345)/(0.5^2) \right )=53`

P=47.62 atm