A) The average translational kinetic energy of the molecules in a gas is given by:
[tex]K = \frac{3}{2}k_B T [/tex]
where
[tex]k_B[/tex] is the Boltzmann's constant
T is the absolute temperature of the gas
In our problem, [tex]T=21.0^{\circ}C +273 = 294 K[/tex], so the average translational kinetic energy of the molecules is
[tex]K= \frac{3}{2}(1.38 \cdot 10^{-23} J/K)(294 K)=6.09 \cdot 10^{-21} J [/tex]
We have 1.2 mol of this gas, and since one mole of ideal gas contains a number of molecules equal to Avogadro number, the total number of molecules in our gas is
[tex]N=n N_A = (1.2 mol)(6.022 \cdot 10^{23} mol^{-1} ) =7.35 \cdot 10^{23}[/tex]
So the total translational kinetic energy of all molecules of the gas is
[tex]K_{tot}= NK = (7.35 \cdot 10^{23})(6.09 \cdot 10^{21} J)=4476 J = 4.5 \cdot 10^3 J[/tex]
B) The kinetic energy of a person is given by:
[tex]K= \frac{1}{2}mv^2 [/tex]
where m is the person's mass and v his velocity. The person has a mass of m=75 kg and its energy is equal to the energy of the gas, [tex]K=4.5 \cdot 10^3 J[/tex], therefore his velocity must be
[tex]v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 (4.5 \cdot 10^3 J)}{75 kg} } =11 m/s [/tex]
