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?

Expert's answer

(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=32⋅1.38⋅10−23 JK⋅423.15 K=8.76⋅10−21 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=3⋅8.3145 JK⋅mol⋅423.15 K4⋅10−3 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=3⋅8.3145 JK⋅mol⋅423.15 K39.9⋅10−3 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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