Answer:
a. E=8.97*10^{-21} J
b. v-He2 = 1.161 km/s
v-Ar2 = 369.74 m/s
Explanation:
(a) The average kinetic energy is given by
[tex]E_k=\frac{1}{2}k_BT[/tex]
where KB is the Boltzman's constant and T is the temperature. For each molecule we have:
[tex]E_k=\frac{3}{2}(1.38066*10^{-23}\frac{J}{K})(433.15K)=8.97*10^{-21}J[/tex]
is the same for both type of molecules because is independent of the mass
(b)
[tex]v_{rms}=\sqrt{\frac{3RT}{M}}[/tex]
where R is the constant of ideal gases, and M is the mass of the molecule. BY replacing for each type of molecule we obtain:
[tex]v_{rms-He_{2}}=\sqrt{\frac{3(8.311434Jmol^{-1}K^{-1})(433.15K)}{0.008kg\ mol^{-1}}}=1161.9\frac{m}{s}\\\\v_{rms-Ar_{2}}=\sqrt{\frac{3(8.311434Jmol^{-1}K^{-1})(433.15K)}{0.079kg\ mol^{-1}}}=369.74\frac{m}{s}[/tex]
v-He2 = 1.161 km/s
v-Ar2 = 369.74 m/s
hope this helps!!
