Question

For a 0.300 mol sample of helium gas in a 0.200 L container at 248K, will the pressure be greater if calculated with the ideal gas law or the van der Waals equation, and by roughly how much? (For He,a=0.0342L2atmmol2,b=0.0237 Lmol)

Answer 1

Answer:

It changes by roughly 1 atm.

Explanation:

Hello!

In this case, since the ideal gas equation differs from the van der Waals' one by the presence of the a and b parameters which correct the assumption of no interactions into the container, they are written as:

$P=\frac{nRT}{V}\\\\P=\frac{RT}{v_m-b}-\frac{a}{v_m^2}$

Thus, the pressure via the ideal gas equation is:

$P=\frac{0.300mol*0.082\frac{atm*L}{mol*K}*248K}{0.200L}=30.5atm$

And the pressure via the van der Waals equation, considering the molar volume (vm=0.200L/0.300L=0.667L/mol) is:

$P=\frac{0.082\frac{atm*L}{mol*K}*248K}{0.667L/mol-0.0237L/mol}-\frac{0.0342atm*L^2/mol^2}{(0.667L/mol)^2}\\\\P=31.6atm-0.0769atm\\\\P=31.5atm$

It means that the pressure change by 1 atm, which is not a significant difference for helium.