Question

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

Answer 1

`\%diff=24.0\%`

Step-by-step 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:

`p=(RT)/(v-b)-(a)/(v^2)`

Thus, we secondly calculate the molar volume as:

`v=(0.8311L)/(9.998mol) =0.083L/mol`

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

`p=(0.08206(atm*L)/(mol*K)*502.7K)/(0.08313L/mol-0.03219L/mol)-(1.345L*atm/mol)/((0.08313L/mol)^2)\\\\p=615.2atm`

And the ideal gas pressure:

`p=(0.08206(atm*L)/(mol*K)*502.7K)/(0.08313L/mol)\\\\p=496.2atm`

Finally, the percent difference:

`\%diff=(|496.2atm-615.2atm|)/(496.2atm) *100\%\\\\\%diff=24.0\%`

Regards!