Homework Solutions

February 3

Chapter 5.

61.    KE(ave) = (3/2)*RT, R + 8.314 J/mol K

        at 273 K KE = 1.5*8.314*273 = 3.40 kJ/mol
        at 546 K KE = 1.5*8.314*546 = 6.81 kJ/mol

62.    U(rms) = (3RT/M)^1/2; where it is easiest to put M in kg/mol

methane at 273 K:  (3*8.314*273/0.01614)^1/2 = 649 m/s

methane at 546 K:  (3*8.314*546/.01614)^1/2 = 918 m/s

nitrogen at 273 K:  (3*8.314*273/.02800)^1/2 = 493 m/s

nitrogen at 546 K:  (3*8.314*546/.02800)^1/2 = 697 m/s

63.    No.  The kinetic energy we calculate is an average kinetic energy.  The velocities we calculate are average values.  The gas molecules travel at different speeds, following the Maxwell-Boltzmann distribution.  Because their mass is constant, mv² will vary as their velocity varies.