Critical Density and Vander Waal's Equation

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SUMMARY

The discussion focuses on calculating the Vander Waals constants 'a' and 'b' using the critical density of 0.555 gm/cc and critical pressure of 48 atm. The correct values derived are a = 5.645 atm lit²/mol² and b = 0.066 lit/mol. Participants highlight the necessity of the molecular weight and critical temperature for accurate calculations, emphasizing that critical density is directly related to constant 'b'. The conversation reveals a common misunderstanding regarding the application of critical density in these calculations.

PREREQUISITES
  • Understanding of Vander Waals equation
  • Knowledge of critical properties in thermodynamics
  • Familiarity with unit conversions (gm/cc to lit/mol)
  • Basic algebra for solving equations with two unknowns
NEXT STEPS
  • Study the derivation of the Vander Waals equation
  • Learn about critical properties and their significance in thermodynamics
  • Explore the relationship between molecular weight and critical constants
  • Investigate common errors in unit conversions in physical chemistry
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Chemists, chemical engineers, and students studying physical chemistry who are interested in the application of the Vander Waals equation and critical properties of substances.

shaiqbashir
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" 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
 
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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.
 
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 I am facing you see is this that how to use this critical density to calculate "b". Now i know two formulae here:

P_{c}=\frac{a}{27b^2}

V_{c}=3b

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 don't you think that we should proceed in this way:

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

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
 
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.
 
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