Question

A 9.439 mol sample of oxygen gas is maintained in a 0.8200 L container at 304.4 K. What is the pressure in atm calculated using the van der Waals' equation for O2 gas under these conditions

Answer 1

Answer:

$P=273.1atm$

Explanation:

Hello!

In this case, since the Van der Waals' equation is used in order to analyze a gas slightly deviated from the ideal condition and is defined as:

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

Whereas a and b for oxygen are 0.0318 L/mol and 1.36 atm*L²/mol² respectively and represent the effective volume and the eventual interactions among the gas molecules. Moreover, the molar volume, vm, is:

$v_m=\frac{0.8200L}{9.439mol}=0.08687L/mol$

Thus, the required pressure turns out:

$P=\frac{0.082\frac{atm*L}{mol*K}*304.4K}{0.08687L/mol-0.0318L/mol}-\frac{1.36\frac{atm*L^2}{mol^2} }{(0.08687L/mol)^2}\\\\P=453.3atm-180.2atm\\\\P=273.1atm$

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