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matrenka [14]
3 years ago
6

My professor gave me two questions to solve using the Van Der Waals Equation. She told us to solve for P and the second one we h

ave to solve for b
She wants an explanation how we got our answers. What confusing is the fact that she wants the answers in units but she never gave us any numbers to plot for the equations unless she wants us to use Si units. She did however say P=atmosphere, b=litters, and T=kelvin


Van der waals equation

(P+n^2a/V^2)(V-nb)=nRT. solve for P

T=kelvin
P=atmosphere
b=litters

profesor wants units, so im guessing she wants us to use Si units

Chemistry
1 answer:
Fed [463]3 years ago
7 0

Answer:

P=atm

b=\frac{L}{mol}

Explanation:

The problem give you the Van Der Waals equation:

(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT

First we are going to solve for P:

(P+\frac{n^{2}a}{V^{2}})=\frac{nRT}{(V-nb)}

P=\frac{nRT}{(V-nb)}-\frac{n^{2}a}{v^{2}}

Then you should know all the units of each term of the equation, that is:

P=atm

n=mol

R=\frac{L.atm}{mol.K}

a=atm\frac{L^{2}}{mol^{2}}

b=\frac{L}{mol}

T=K

V=L

where atm=atmosphere, L=litters, K=kelvin

Now, you should replace the units in the equation for each value:

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}}

Then you should multiply and eliminate the same units which they are dividing each other (Please see the photo below), so you have:

P=\frac{L.atm}{L-L}-atm

Then operate the fraction subtraction:

P=P=\frac{L.atm-L.atm}{L}

P=\frac{L.atm}{L}

And finally you can find the answer:

P=atm

Now solving for b:

(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT

(V-nb)=\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}

nb=V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}

b=\frac{V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}}{n}

Replacing units:

b=\frac{L-\frac{(mol).(\frac{L.atm}{mol.K}).K}{(atm+\frac{mol^{2}.\frac{atm.L^{2}}{mol^{2}}}{L^{2}})}}{mol}

Multiplying and dividing units,(please see the second photo below), we have:

b=\frac{L-\frac{L.atm}{atm}}{mol}

b=\frac{L-L}{mol}

b=\frac{L}{mol}

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