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12     CHAPTER 1 Fundamental Concepts of Thermodynamics

                                            In Example Problem 1.4, a comparison is made of the molar volume for N 2  calcu-
                                         lated at low and high pressures, using the ideal gas and van der Waals equations of state.



                                          EXAMPLE PROBLEM 1.4

                                          a. Calculate the pressure exerted by N 2  at 300. K for molar volumes of 250. L mol -1
                                             and 0.100 L mol -1  using the ideal gas and the van der Waals equations of state.
                                                                                           6
                                             The values of parameters a and b for N 2  are 1.370 bar dm  mol -2  and
                                                     3
                                             0.0387 dm  mol -1 , respectively.
                                          b. Compare the results of your calculations at the two pressures. If P calculated
                                             using the van der Waals equation of state is greater than those calculated with the
                                             ideal gas law, we can conclude that the repulsive interaction of the N 2  molecules
                                             outweighs the attractive interaction for the calculated value of the density. A
                                             similar statement can be made regarding the attractive interaction. Is the attrac-
                                             tive or repulsive interaction greater for N 2  at 300. K and V m  = 0.100 L ?

                                         Solution
                                          a. The pressures calculated from the ideal gas equation of state are
                                                                                -1
                                                                      -2
                                              nRT    1mol * 8.314 * 10  L bar mol K -1  * 300. K
                                          P =      =                                           = 9.98 * 10 -2  bar
                                               V                       250. L
                                                                                -1
                                                                      -2
                                              nRT    1 mol * 8.314 * 10  L bar mol K -1  * 300. K
                                          P =      =                                           = 249 bar
                                               V                      0.100 L
                                             The pressures calculated from the van der Waals equation of state are
                                                                    2
                                                           nRT     n a
                                                      P =        -
                                                          V - nb    V 2
                                                                           -2
                                                                                     -1
                                                          1mol * 8.314 * 10  L bar mol K -1  * 300. K
                                                         =
                                                                                       3
                                                               250. L - 1mol * 0.0387 dm  mol -1
                                                                  2
                                                                                6
                                                            (1mol) * 1.370 bar dm  mol -2
                                                            -
                                                                     (250. L) 2
                                                         = 9.98 * 10 -2  bar
                                                                                     -1
                                                                           -2
                                                          1mol * 8.314 * 10  L bar mol K -1  * 300 K
                                                       P =
                                                                                        3
                                                              0.100 L - 1mol * 0.0387 dm  mol -1
                                                                                6
                                                                  2
                                                            (1mol) * 1.370 bar dm  mol -2
                                                            -
                                                                     (0.100 L) 2
                                                         = 270. bar
                                          b. Note that the result is identical with that for the ideal gas law for V m  = 250. L ,
                                             and that the result calculated for V m  = 0.100 L  deviates from the ideal gas law
                                             result. Because P real  7 P ideal , we conclude that the repulsive interaction is more
                                             important than the attractive interaction for this specific value of molar volume
                                             and temperature.



        Vocabulary


        absolute temperature scale         Celsius scale                      elastic collision
        adiabatic                          centigrade scale                   equation of state
        Boltzmann constant                 closed system                      equilibrium
        boundary                           diathermal                         extensive variable
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