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

`P=273.1atm`

Step-by-step 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=(RT)/(v_m-b)-(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=(0.8200L)/(9.439mol)=0.08687L/mol`

Thus, the required pressure turns out:

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

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