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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

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Mar 2, 2018

Answer:

#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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