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The average energy of molecules in a sample of oxygen gas at 300 K are 6.21xx10^(-21)J. The corresponding values at 600 K are |
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Answer» `12.12xx10^(-21)J` `KE_(avg)=(3)/(2)k_(B)Tor((KE_(avg))_(1))/((KE_(avg))_(2))=(T_(1))/(T_(2))` Here, `(KE_(avg))_(1)=6.21xx10^(-21)J,T_(1)=300K`, `T_(2)=600K,(KE_(avg))_(2)=?` `therefore (6.21xx10^(-21))/((KE_(avg))_(2))=(300)/(600)=(1)/(2)or(KE_(avg))_(2)=12.42xx10^(-21)J` |
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