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10     CHAPTER 1 Fundamental Concepts of Thermodynamics

                                         Solution
                                         The number of moles of He, Ne, and Xe is given by
                                                     PV              1.50bar * 2.00L
                                                   =     =
                                                n He                 -2      -1    -1         = 0.121 mol
                                                     RT    8.314 * 10  L bar K  mol   * 298 K
                                                     PV              2.50bar * 3.00 L
                                                   =     =
                                                n Ne                 -2      -1    -1         = 0.303 mol
                                                     RT    8.314 * 10  L bar K  mol   * 298 K
                                                     PV              1.00bar * 1.00 L
                                                n Xe  =  =                                    = 0.0403 mol
                                                                             -1
                                                                     -2
                                                     RT    8.314 * 10  L bar K  mol -1  * 298 K
                                                  n = n He  + n Ne  + n Xe  = 0.464
                                         The mole fractions are
                                                                     n He  0.121
                                                               x He  =   =       = 0.261
                                                                      n    0.464
                                                                     n Ne  0.303
                                                               x Ne  =   =       = 0.653
                                                                      n    0.464
                                                                     n Xe  0.0403
                                                               x Xe  =   =       = 0.0860
                                                                      n    0.464
                                         The total pressure is given by
                                                        (n He  + n Ne  + n )RT
                                                                     Xe
                                                    P =
                                                                 V
                                                                                     -1
                                                                             -2
                                                        0.464 mol * 8.314 * 10  L bar K  mol -1  * 298 K
                                                       =
                                                                            6.00 L
                                                       = 1.92 bar
                                         The partial pressures are given by
                                                         P He  = x P = 0.261 * 1.92 bar = 0.501 bar
                                                                He
                                                         P Ne  = x P = 0.653 * 1.92 bar = 1.25 bar
                                                                Ne
                                                         P Xe  = x P = 0.0860 * 1.92 bar = 0.165 bar
                                                                Xe





                                         1.5 A Brief Introduction to Real Gases
           273.6
                                         The ideal gas law provides a first look at the usefulness of describing a system in
                                         terms of macroscopic parameters. However, we should also emphasize the downside
           273.4                 Ar
                                         of not taking the microscopic nature of the system into account. For example, the ideal
                                   N 2   gas law only holds for gases at low densities. In practice, deviations from the ideal gas
          T /K  273.2                    law that occur for real gases must be taken into account in such applications as a gas
                                                                                                           for a
                                         thermometer. If data were obtained from a gas thermometer using He, Ar, and N
                                                                                                          2
                                         temperature very near the temperature at which the gas condenses to form a liquid,
                                  He
                                         they would exhibit the behavior shown in Figure 1.9. We see that the temperature only
           273.0
                                         becomes independent of  P and of the gas used in the thermometer if the data are
                     2   4   6   8  10   extrapolated to zero pressure. It is in this limit that the gas thermometer provides a
                         10  4  P/Pa
                                         measure of the thermodynamic temperature. In practice, gas-independent T values are
           272.8                                            '
                                         only obtained below P  0.01 bar .
        FIGURE 1.9                          For most applications, calculations based on the ideal gas law are valid to much
        The temperature measured in a gas ther-  higher pressures. Real gases will be discussed in detail in Chapter 7. However, because
        mometer is independent of the gas used  we need to take nonideal gas behavior into account in Chapters 2 through 6, we intro-
        only in the limit that P : 0 .   duce an equation of state that is valid to higher densities in this section.
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