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.

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Netta00 Netta00 · Answer 1 · 2019-10-28T12:22:19+01:00

Answer:

Ideal ,P=49.52 atm

Real ,P=47.62 atm

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

[tex]\left ( P+\dfrac{an^2}{v^2} \right )\left ( v-nb \right )=nRT[/tex]

Now by putting the values

[tex]\left ( P+\dfrac{1.345\times 1^2}{0.5^2} \right )\left ( 0.5-1\times 0.03219 \right )=1\times 0.0821\times 302[/tex]

[tex]\left ( P+\dfrac{1.345}{0.5^2} \right )=53[/tex]

P=47.62 atm