Solutions to Week 13 Assignment

1. (a) T_C = (T - 273.15K)/(1K/°C) = (77K - 273.15K)/(1K/°C) = -196C
   (b) T_F = (1.8°F/°C)T_C +32°F = (1.8°F/°C)(196.15°C) + 32°F = -321°F.
2. From a proportion with DL_Pb  and DL_glass (which are set equal) to solve for T_glass when T_Pb = 50.0°C.
   DL_Pb/DL_glass = L₀a_PbDT_Pb/L₀a_glassDT_glass
                      1 = a_PbDT_Pb/a_glassDT_glass
            DT_glass = (a_Pb/a_glass)DT_Pb
               T_glass = (29 x 10^-6K^-1)/(9.4 x 10^-6K^-1)(50.0°C -20.0°C) + 20.0°C = 113°C.
3. (a) N/V = nN_A/V = (1.00 mol)(6.02 x 10²³ mol^-1)/(0.0224m³) = 2.69 x 10²⁵ m³
   (b) Assume that each atom is at the center of a sphere of radius r. The volume of the sphere is
        V/N = 1/(N/V) = 1/(2.69 x 10²⁵ m³) = 3.72 x 10^-26 m³ per atom
        Then V/N = 4/3pr³» 4r³
        The distance separating atoms is approximately the diameter of the spheres.
        Solve for d = 2r.
        d = 2r = 2 (V/4N)^1/3 = 2(3.72 x 10^-26 m³/4)^1/3 = 4nm.
   (c) m = (1.00 mol)[2(14.00674g/mol)] = 28.0g
           r = m/V = (28.0g/0.0224m³)(1kg/10³g) = 1.25kg/m³
4. The volume and moles of the gas are constant.
   P_f/T_f = P_i/T_i
        P_f = (T_i /T_f )P_i
            = [(70.0K + 237.15K)/(20.0K + 237.15K)](115kPa) = 135kPa.
5. N = PV/kT, and V, k and T are constant, so 
   DN = VDP/kT  = [(1.0m³)(15.0atm - 20.0atm)(1.013 x 10⁵ Pa/m)]]/(1.38 x 10^-23 J/K) = -1.3 x 10²⁶
   1.3 x 10²⁶ air molecules were released.
6. v_rms = Ö3kT/m = Ö[3(1.38 x 10^-23J/K)(273.15K + 27K)]/(1.38 x 10^-17Kg) = 3.00cm/s.