Question

Calculate P versus molar volume behavior of water at three different temperatures: 300 K, 400 K and 500 K using  vander Waals equation of state.  Your results should show both extrema. 

Using the diagrams you generated in part 1, determine the equilibrium pressure:  the pressure at which the areas above and below the isobar are identical.  Compare your results with the data reported in the steam tables. 

Answer 1



m
2 11
1
8.206 10 L atmK mol 300K
3.00atm
8.206Lmol
RT V P
be sure to use the value of R with the correct units!
Substitute the molar volume from part (a) into the molar volume on the right hand side of the above
equation (the term in red) using a = 3.64 L2 atm mol-2 and b = 4.267  10-2 L mol-1:
    
 
 




     
  
 
2 11
2 1 -1
m 2 2
2 1
8.206 10 L atmK mol 300K
1 4.267 10 Lmol 8.103 L mol
3.64L atmmol 3.
8.206Lmol
00atm
V
Substitution of this result into the expression for Vm gives
Substituting Vm(2) for Vm gives
    
 
 




     
  
 
2 11
2 1 -1
m 2 2
2 1
8.206 10 L atmK mol 300K
2 4.267 10 Lmol 8.100 L mol
3.64L atmmol 3.
8.103Lmol
00atm
V
Substituting Vm(3) for Vm gives
    
 
 




     
  
 
2 11
2 1 -1
m 2 2
2 1
8.206 10 L atmK mol 300K
3 4.267 10 Lmol 8.099 L mol
3.64L atmmol 3.
8.100Lmol
00atm
V
Since Vm(4) = Vm(3), the cycle of approximation has converged and can be terminated. (Note that no
matter how you solve for Vm, successive approximations will always give the same result, but may take
more iterations to converge.)
    
 
 




     
  
 
2 11
2 1 -1
m 2 2
2 1
8.206 10 L atmK mol 300K
4 4.267 10 Lmol 8.099 L mol
3.64L atmmol 3.
8.099Lmol
00atm
V
  

  

 


 

2
m m
2
2
m m
2
m
m
m
m
RT a P
V b V
a RT P
V V b
RT V b
b
V
V
a P
RT V
a P
2.3 (a) Calculate the molar volume of an ideal gas at 300 K and 3.00 atm. (b) Use the van der Waals equation of
state to determine the molar volume of CO2 gas at the same temperature and pressure. For (b), the van
der Waals equation cannot be solved for Vm explicitly—any expression for Vm will include one or more molar
volume terms on the right hand side. You can use the answer from (a) as a first approximation for the
molar volume term on the right hand side, then use successive approximations to obtain a numerical
solution. The van der Waals coefficients, a and b, can be found in Table 2.2.
**Solution**
a. For an ideal gas, the molar volume is
b. Rearrange the van der Waals equation