Question

A 9.380 mol sample of nitrogen gas is maintained in a 0.8198 L container at 301.8 K. What is the pressure in atm calculated using the van der Waals' equation for N2 gas under these conditions? For N2, a = 1.390 L2atm/mol2 and b = 3.910×10-2 L/mol. __ atm

Answer 1

`P=331.35"atm"`

Explanation:

The Van der Waals' equation states:

`(P+(an^2)/V^2)(V-nb)=nRT`

To solve for `P`, we can write:

`(P+(an^2)/V^2)=(nRT)/(V-nb)`

`P=(nRT)/(V-nb)-(an^2)/V^2`

Here,

`n=9.38"mol"`

`V=0.8198"L"`

`T=301.8"K"`

`a=1.39"L"^2"atm mol"^-2`

`b=3.91xx10^-2"L mol"^-1 "atm"^-1`

`R=0.0821"L atm K"^-1"mol"^-1`

Inputting:

`P=(9.38*0.0821*301.8)/(0.8198-9.38*3.91xx10^-2)-(1.39*9.38^2)/0.8198^2`

`P=331.35"atm"`

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