A) \[{{({{v}_{rms}})}_{1}}<{{({{v}_{rms}})}_{2}}<{{({{v}_{rms}})}_{3}}\] and \[{{(\bar{K})}_{1}}={{(\bar{K})}_{2}}={{(\bar{K})}_{3}}\]
B) \[{{({{v}_{rms}})}_{1}}={{({{v}_{rms}})}_{2}}={{({{v}_{rms}})}_{3}}\]and \[{{(\bar{K})}_{1}}={{(\bar{K})}_{2}}>{{(\bar{K})}_{3}}\]
C) \[{{({{v}_{rms}})}_{1}}>{{({{v}_{rms}})}_{2}}>{{({{v}_{rms}})}_{3}}\] and \[{{(\bar{K})}_{1}}<{{(\bar{K})}_{2}}>{{(\bar{K})}_{3}}\]
D) \[{{({{v}_{rms}})}_{1}}>{{({{v}_{rms}})}_{2}}>{{({{v}_{rms}})}_{3}}\]and \[{{(\bar{K})}_{1}}<{{(\bar{K})}_{2}}<{{(\bar{K})}_{3}}\]
Correct Answer: A
Solution :
[a] \[{{v}_{rms}}\propto \frac{1}{\sqrt{M}}\Rightarrow {{({{v}_{rms}})}_{1}}<{{({{v}_{rms}})}_{2}}<{{({{v}_{rms}})}_{3}}\] Also in mixture temperature of each gas will be same, hence kinetic energy also remains same.You need to login to perform this action.
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