A) \[{{\left( {{V}_{rms}} \right)}_{1}}<\text{ }{{\left( {{V}_{rms}} \right)}_{2}}<\text{ }{{\left( {{V}_{rms}} \right)}_{3}}\] and \[{{(\overline{K})}_{1}}={{(\overline{K})}_{2}}=({{\overline{K}}_{3}})\]
B) \[{{\left( {{V}_{rms}} \right)}_{1}}=\text{ }{{\left( {{V}_{rms}} \right)}_{2}}=\text{ }{{\left( {{V}_{rms}} \right)}_{3}}\] and \[{{(\overline{K})}_{1}}={{(\overline{K})}_{2}}>{{(\overline{K})}_{3}}\]
C) \[{{\left( {{V}_{rms}} \right)}_{1}}>{{\left( {{V}_{rms}} \right)}_{2}}>{{\left( {{V}_{rms}} \right)}_{3}}\] and \[{{(\overline{K})}_{1}}<{{(\overline{K})}_{2}}>({{\overline{K}}_{3}})\]
D) \[{{\left( {{V}_{rms}} \right)}_{1}}>{{\left( {{V}_{rms}} \right)}_{2}}>{{\left( {{V}_{rms}} \right)}_{3}}\] and \[{{(\overline{K})}_{1}}<{{(\overline{K})}_{2}}<{{(\overline{K})}_{3}}\]
Correct Answer: A
Solution :
[a] \[{{v}_{rms}}\frac{1}{\sqrt{M}}\Rightarrow {{({{v}_{rms}})}_{1}}<{{({{v}_{rms}})}_{2}}<{{({{v}_{rms}})}_{3}}\] also in mixture temperature of each gas will be same, hence kinetic energy also remains same.
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