Question

A cylinder contains a mixture of helium and argon gas in equilibrium at 160°C.

Answer 1

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

$E_k=\frac{1}{2}k_BT$

where KB is the Boltzman's constant and T is the temperature. For each molecule we have:

$E_k=\frac{3}{2}(1.38066*10^{-23}\frac{J}{K})(433.15K)=8.97*10^{-21}J$

is the same for both type of molecules because is independent of the mass

(b)

$v_{rms}=\sqrt{\frac{3RT}{M}}$

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:

$v_{rms-He_{2}}=\sqrt{\frac{3(8.311434Jmol^{-1}K^{-1})(433.15K)}{0.008kg\ mol^{-1}}}=1161.9\frac{m}{s}\\\\v_{rms-Ar_{2}}=\sqrt{\frac{3(8.311434Jmol^{-1}K^{-1})(433.15K)}{0.079kg\ mol^{-1}}}=369.74\frac{m}{s}$

v-He2 = 1.161 km/s

v-Ar2 = 369.74 m/s

hope this helps!!