Question

Pretty desperate here. Please help me.

Answer 1

Express the van der Waals equation of state as a virial expansion in powers of n/V, keeping terms up to the third power of (n/V). The result should have the form:
\[\frac{pV}{nRT}=1+B(\frac{n}{v})+C(\frac{n}{v})^{2}+D(\frac{n}{v})^{3}+...\]
Express the B, C, D virial coefficients in terms of the van der Waals parameters a and b.

Hint: Use the Taylor-series expansion 1/(1-x) = 1 + x + x2 + ..., for |x|< 1.

Answer 2

@agent0smith

Answer 3

i really don't understand this at all
should be:
\[\frac{nRT}{V}=1+B(\frac{n}{v})+C(\frac{n}{v})^{2}+D(\frac{n}{v})^{3}+...\rightarrow\]
\[\frac{RT}{V_{m}}=1+B(\frac{n}{V_{m}})+C(\frac{n}{V_{m}})^{2}+D(\frac{n}{V_{m} })^{3}+...\]

Answer 4

so:
\[p=\frac{RT(1-\frac{b}{V_{m} })^{-1}}{V_{m} }-\frac{a}{V_{m} ^{2}}\]

Answer 5

then you're supposed to do this?:
\[p=\frac{RT}{V_{m} }((-1\frac{b}{V_{m} })^{-1} -\frac{a}{RTV_{m} })\]

Answer 6

@jim_thompson5910 @mathmale

Answer 7

then i was told to do a taylor expansion that is supposed to look like this:
\[p=\frac{RT}{V_{m} }(1+(b-\frac{a}{RT })(\frac{1}{V_{m}}) +b^{2}\frac{1}{V_{m}^{2} })+...\]
I don't know what I'm doing =(

Answer 8

\[B=b-\frac{a}{RT}\]
\[C=b^{2}\]
what the heck is D

Answer 9

"pretty desperate here" and so many people on your post XD thought you were looking for a bf ROFL x'D

Answer 10

Ok I'm trying to get this straight, which is understanding what all these variables mean, is n/V the density here? I mean from what I see we may need to think about the thermodynamic identity which requires derivatives. And the a and b are correctional terms correct (constants)?

Sorry I'm trying to understand this as well, because I don't know why we're using a taylor series yet, I can see using other methods but my chemistry isn't very good yet as most of it is self taught ha.

Answer 11

molar volume I believe
yes a and b are constants. I've been studying chemistry for years, but this is the first time I've actually had to dive deep into physical chemistry

Answer 12

Haha yeah it's a bit tricky, but I've only done high school chemistry and am finally taking general chemistry now but let me see if I can get the gist of this...

Answer 13

i was pretty much given the first two but i dont know how to come about the last one. its very frustrating

Answer 14

Ok well I'm going to tag an actual chemist and let him know about this problem he would find it interesting I think @Kainui

Answer 15

i hope so

Answer 16

Ideal gas equation :
\[\dfrac{pV}{nRT}=1\]

I think that ratio changes when the density of the gas is not low. The given expansion seems to model real gases...

Answer 17

i'm using van der waals because this case is for a non ideal gas with different conditions.

Answer 18

Is this the starting van der Waals equation
\[(p+a/V^2)(V-b)= RT\]

?

Answer 19

Yeah that looks like an equation of state

Answer 20

Looks it was then rearranged as :
\[p=\frac{RT\color{red}{(1-\frac{b}{V_{m} })^{-1}}}{V_{m} }-\frac{a}{V_{m} ^{2}}\]

Answer 21

Plugging in \[\color{red}{(1-\frac{b}{V_{m} })^{-1}= 1 + \frac{b}{V_{m} } + (\frac{b}{V_{m} })^2 + \cdots }\]

in the earlier equation gives
\[p=\frac{RT\color{red}{(1 + \frac{b}{V_{m} } + (\frac{b}{V_{m} })^2 + \cdots )}}{V_{m} }-\frac{a}{V_{m} ^{2}}\]

Answer 22

Nicee

Answer 23

I don't see how it is possible to change the variable b/V to n/V in all the powers hmm

Answer 24

They want the expansion to have powers of n/V

Answer 25

What if we started with \[\left( P+\frac{ an^2 }{ V^2 } \right)(V-nb)=nRT\] I mean it's just another way of writing the van der Waals equation of state I think we can get it maybe

Answer 26

Ohkay, \(V_m = V/n\) is the molar mass is it ?

Answer 27

*molar volume

Answer 28

Yeah she mentioned that above as well

Answer 29

@ganeshie8 i can see what you did for the expansion but I still dont know how to get the D coefficient. the first part was given to e but i cant find D