1.

The average translational energy and the rms speed of molecules in a sample of oxygen gas at `300K` are `6.21xx10^(-21)J` and `484m//s`, respectively. The corresponding values at `600K` are nearly (assuming ideal gas behaviour)A. `12.42xx10^(-21)J, 968 m//s`B. `8.78xx10^(-21)J, 684 m//s`C. `6.21xx10^(-21)J, 968m//s`D. `12.42xx10^(-21)J, 684m//s`

Answer» Correct Answer - D
The formula for average kinetic energy is
`bar((KE))=(3)/(2)KT`
`:. (bar((KE))_(600K))/(bar((KE))_(300K))=(600)/(300)`
`implies bar((KE))_(600K)=2xx6.21xx10^(-21)J=12.42xx10^(-21)J`
Also the formula for rms velocity is
`C_("rms")=sqrt((3KT)/(m))`
`:. ((C_("rms"))_(600K))/((C_("rms"))_(300K))=sqrt((600)/(300))`
`implies (C_("rms"))_(600K)=sqrt(2)xx484=684m//s`.


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