Answer:
P=atm
[tex]b=\frac{L}{mol}[/tex]
Explanation:
The problem give you the Van Der Waals equation:
[tex](P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT[/tex]
First we are going to solve for P:
[tex](P+\frac{n^{2}a}{V^{2}})=\frac{nRT}{(V-nb)}[/tex]
[tex]P=\frac{nRT}{(V-nb)}-\frac{n^{2}a}{v^{2}}[/tex]
Then you should know all the units of each term of the equation, that is:
[tex]P=atm[/tex]
[tex]n=mol[/tex]
[tex]R=\frac{L.atm}{mol.K}[/tex]
[tex]a=atm\frac{L^{2}}{mol^{2}}[/tex]
[tex]b=\frac{L}{mol}[/tex]
[tex]T=K[/tex]
[tex]V=L[/tex]
where atm=atmosphere, L=litters, K=kelvin
Now, you should replace the units in the equation for each value:
[tex]P=\frac{(mol)(\frac{L.atm}{mol.K})(K)}{L-(mol)(\frac{L}{mol})}-\frac{(mol^{2})(\frac{atm.L^{2}}{mol^{2}})}{L^{2}}[/tex]
Then you should multiply and eliminate the same units which they are dividing each other (Please see the photo below), so you have:
[tex]P=\frac{L.atm}{L-L}-atm[/tex]
Then operate the fraction subtraction:
P=[tex]P=\frac{L.atm-L.atm}{L}[/tex]
[tex]P=\frac{L.atm}{L}[/tex]
And finally you can find the answer:
P=atm
Now solving for b:
[tex](P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT[/tex]
[tex](V-nb)=\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}[/tex]
[tex]nb=V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}[/tex]
[tex]b=\frac{V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}}{n}[/tex]
Replacing units:
[tex]b=\frac{L-\frac{(mol).(\frac{L.atm}{mol.K}).K}{(atm+\frac{mol^{2}.\frac{atm.L^{2}}{mol^{2}}}{L^{2}})}}{mol}[/tex]
Multiplying and dividing units,(please see the second photo below), we have:
[tex]b=\frac{L-\frac{L.atm}{atm}}{mol}[/tex]
[tex]b=\frac{L-L}{mol}[/tex]
[tex]b=\frac{L}{mol}[/tex]
