Page 254 - Schaum's Outline of Theory and Problems of Applied Physics
P. 254

CHAP. 20]                        KINETIC THEORY OF MATTER                             239



         p = 1 atm, and V = (n)(22.4 L/mol) so that
                                      pV    (n)(1 atm)(22.4 L/mol)
                                         =                     = nR
                                      T            273 K
        where R, the universal gas constant, has the value


                                          R = 0.0821 atm·L/(mol·K)
        In SI units, in which p is in newtons per square meter and V is in cubic meters,

                                             R = 8.31 J/(mol·K)

        The complete ideal gas law is usually written in the form

                                                pV = nRT


        SOLVED PROBLEM 20.13
                                                                                               ◦
              (a) What volume does1gof ammonia (NH 3 ) occupy at STP? (b) What volume does it occupy at 100 C
              and a pressure of 1.2 atm?
              (a) The molecular mass of NH 3 is

                                         1(N) = (1)(14.01) u = 14.01 u
                                         3(H) = (3)(1.008) u = 3.02 u
                                                          17.03 u = 17.03 g/mol

                  so the number of moles in1gofNH 3 is

                                                mass of NH 3       1g
                               Moles of NH 3 =                =          = 0.0587 mol
                                            molecular mass of NH 3  17.03 g/mol
                      The volume at STP is therefore

                         Volume of NH 3 = (moles of NH 3 )(molar volume) = (0.0587 mol)(22.4 L/mol) = 1.32 L

              (b) From the ideal gas law,

                                           p 1 V 1  p 2 V 2        p 1 V 1 T 2
                                                =        or   V 2 =
                                            T 1    T 2              p 2 T 1
                                              ◦
                                                                                 ◦
                  Here p 1 = 1 atm, V 1 = 1.32 L, T 1 = 0 C = 273 K and p 2 = 1.2 atm, V 2 =?, T 2 = 100 C = 373 K.
                  Hence
                                               (1 atm)(1.32 L)(373 K)
                                           V 2 =                 = 1.50 L
                                                  (1.2 atm)(273 K)

        SOLVED PROBLEM 20.14
                                                                 ◦
              What is the mass of 40 L of uranium hexafluoride (UF 6 )at500 C and 4 atm of pressure?
                  The most direct way to solve this problem is to use the ideal gas law to find the number of moles of UF 6 in the
              sample. Since pV = nRT and T = 500 C = 773 K, we have
                                           ◦
                                       pV          (4 atm)(40 L)
                                   n =    =                         = 2.52 mol
                                       RT   [0.0821 atm·L/(mol·K)](773 K)
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