Question #160589
A cylinder contains a mixture of helium and argon gas in equilibrium at 150 oC.
(a) What is the average kinetic energy for each type of gas molecule?
(b) What is the rms speed for each type of gas molecule?
1
Expert's answer
2021-02-07T19:17:34-0500

(a) Since the average kinetic energy depends only on the temperature of the gas and mixture of helium and argon gases in equilibrium at 150 C150\ ^{\circ}C, both gas molecules have the same average kinetic energy:


KEavg=32kBT,KE_{avg}=\dfrac{3}{2}k_BT,KEavg=321.381023 JK423.15 K=8.761021 J.KE_{avg}=\dfrac{3}{2}\cdot1.38\cdot10^{-23}\ \dfrac{J}{K}\cdot423.15\ K=8.76\cdot10^{-21}\ J.

(b) Let's find the rms speed of helium gas molecule:


vrms,He=3RTMHe,v_{rms,He}=\sqrt{\dfrac{3RT}{M_{He}}},vrms,He=38.3145 JKmol423.15 K4103 kgmol=1624 ms.v_{rms,He}=\sqrt{\dfrac{3\cdot8.3145\ \dfrac{J}{K\cdot mol}\cdot423.15\ K}{4\cdot10^{-3}\ \dfrac{kg}{mol}}}=1624\ \dfrac{m}{s}.

Let's find the rms speed of argon gas molecule:


vrms,Ar=3RTMAr,v_{rms,Ar}=\sqrt{\dfrac{3RT}{M_{Ar}}},vrms,Ar=38.3145 JKmol423.15 K39.9103 kgmol=514 ms.v_{rms,Ar}=\sqrt{\dfrac{3\cdot8.3145\ \dfrac{J}{K\cdot mol}\cdot423.15\ K}{39.9\cdot10^{-3}\ \dfrac{kg}{mol}}}=514\ \dfrac{m}{s}.

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