According to the ideal gas law, a 9.998 mol sample of argon gas in a 0.8311 L container at 502.7 K should exert a pressure of 496.2
atm. What is the percent difference between the pressure calculated using the van der Waals' equation and the ideal pressure? For Ar
gas, a = 1.345 L’atm/mol? and b = 3.219x10-2 L/mol.
Pideal – Puan der Waals |
Percent difference
x 100

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sebassandin sebassandin · Answer 1 · 2021-07-23T02:24:51+02:00

Answer:

[tex]\%diff=24.0\%[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out firstly necessary for us to set up the van der Waals' equation as shown below:

[tex]p=\frac{RT}{v-b}-\frac{a}{v^2}[/tex]

Thus, we secondly calculate the molar volume as:

[tex]v=\frac{0.8311L}{9.998mol} =0.083L/mol[/tex]

Then, we plug in the entire variables in the vdW equation to get such pressure:

[tex]p=\frac{0.08206\frac{atm*L}{mol*K}*502.7K}{0.08313L/mol-0.03219L/mol}-\frac{1.345L*atm/mol}{(0.08313L/mol)^2}\\\\p=615.2atm[/tex]

And the ideal gas pressure:

[tex]p=\frac{0.08206\frac{atm*L}{mol*K}*502.7K}{0.08313L/mol}\\\\p=496.2atm[/tex]

Finally, the percent difference:

[tex]\%diff=\frac{|496.2atm-615.2atm|}{496.2atm} *100\%\\\\\%diff=24.0\%[/tex]

Regards!