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)

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:

[tex]P=\frac{nRT}{V}\\\\P=\frac{RT}{v_m-b}-\frac{a}{v_m^2}[/tex]

Thus, the pressure via the ideal gas equation is:

[tex]P=\frac{0.300mol*0.082\frac{atm*L}{mol*K}*248K}{0.200L}=30.5atm[/tex]

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

[tex]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[/tex]

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

pstnonsonjoku pstnonsonjoku · Answer 2 · 2021-10-28T14:46:42+02:00

The difference in pressure calculated by the two methods is 84 atm.

The ideal gas equation is given by

PV =nRT

From the data given in the question;

P = ?

V = 0.200 L

n = 0.300 mol

T = 248K

R = 0.082 atmLK-1Mol-1

P = nRT/V

P = 0.300 mol × 0.082 atmLK-1Mol-1 × 248K/0.200 L

P = 30.5 atm

From Van der Waals equation;

P = RT/V - b - a/V^2

P = (0.082 × 248/0.200 - 0.0237) - (0.0342/ 0.200^2)

P = 114.5 atm

The difference in pressure calculated by the two methods is;

114.5 atm - 30.5 atm = 84 atm

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