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Thermodynamics ?

  • Thread starter mkbh_10
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  • #1
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1. Homework Statement

calculate critical temp. for a gas obeying vander waal eqn of state . Give a = .00874 atm (cm)^6 & b = .0023 (cm)^3 for 1 (cm)^3 of gas at STP .

2. Homework Equations



3. The Attempt at a Solution


unit of a is N (m)^4 (mol)^-2 , how do i get this mol term ?

(cm)^6 = (10)^-12 (m)^6 , pressure at STP is 1 atm = 1.013*(10)^5 N/m(sqr) ,

combining these two this equals

1.013 (10)^-7 N (m)^4 , to get the per mol term , i use the eqn of state PV=nRT where

p = 1 atm , V= 22.4 ltrs = 22.4 *(10)^-3 (m)^3 , R= 8.31 J/k/mol T = 273 K

I am stuck after this , is the conversion of volume correct ?
 

Answers and Replies

  • #2
Hootenanny
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What is the equation of state for a Van der Waal's gas?
 
  • #3
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[p+ a/(V)^2](V-b) = nRT
 
  • #4
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Critical temp. = 8a/27Rb
 
  • #5
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I use PV=nrt to get a & b for 1 mole of a gas but i am stuck at volume conversion
 
  • #6
Hootenanny
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Then all you need to do is sub the values in for a and b, the volume doesn't come into it.
 
  • #7
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But R has unit in N/m(sqr) 1/k 1/mol , this mol unit will come into play if i put the values of a & b directly , have to remove it so i will have to change
 
  • #8
Hootenanny
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But R has unit in N/m(sqr) 1/k 1/mol , this mol unit will come into play if i put the values of a & b directly , have to remove it so i will have to change
Ahh true (sorry, missed that :redface:). However, you need to use Van der Waal's equation to find the number of moles that you are dealing with, not the ideal gas equation. After you have found the number of moles you are dealing with simply multiply your expression for the critical temperature by that number.
 
  • #9
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I am still not able to get the correct value
 

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