Critical Density and Vander Waal's Equation

  • #1
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Main Question or Discussion Point

" Critical Density of a substance having molecular weight is 0.555 gm/cc and the critical pressure is 48 atm. Calculate the Vander Waals constants 'a' and 'b'. "

Ans: a= 5.645 atm lit^2/mol^2
b= 0.066 lit/mol

My dear friendS! it looks like to me that the data is incomplete here as there is not molecular mass or critical temperature given there. Can you help me in solving this prob. I shall be very thankful to you for this act of kindness.

Thanks and Good Bye
 

Answers and Replies

  • #2
Gokul43201
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Shaiq, the data is complete. There are two unknowns 'a', and 'b', and both the critical pressure and critical density depend only on these two constants (in fact, the critical density is related only to 'b').

Two equations (do you know them?), two unknowns.
 
  • #3
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how to use critical density there!!!!

Hi Gokul!!!

Well! my dear friend! thanks a lot for your precious help. But here's a little problem which im facing you see is this that how to use this critical density to calculate "b". Now i know two formulae here:

[tex]P_{c}=\frac{a}{27b^2}[/tex]

[tex]V_{c}=3b[/tex]

Now my dear friend!! i know that density is mass/vol. Now can i write critical density as critical mass/critical volume. If supposing that this is the formula then dont you think that we should proceed in this way:

[tex]V_{c}=3b[/tex]
[tex]\frac{m_{c}}{V_{c}}=\frac{m_{c}}{3b}[/tex]
[tex]0.555gm/cc=\frac{0.555gm}{3b}[/tex]
[tex]b=\frac{0.555gm}{3\times0.555gm/cc}[/tex]

now this yield in a value of b=0.333 which is not correct as in the answers, the correct value of b is b=0.066 lit/mol

what should i do now?? please help me as soon as possible
 
  • #4
Gokul43201
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I'm sorry, I didn't look at the units carefully before. I apologize.

shaiqbashir said:
" Critical Density of a substance having molecular weight <number missing here> is 0.555 gm/cc and the critical pressure is 48 atm. Calculate the Vander Waals constants 'a' and 'b'. "
You ARE correct. Actually, I think you may have copied the question down incompletely. It promises a molecular weight but does not give you the number. Yes, you need the MW.
 
Last edited:

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