The root mean square velocity of gas molecules at `27^(@)C` is `1365ms^(-1)` .The gas is
The root mean square velocity of gas molecules at `27^(@)C` is `1365ms^(-1)` .The gas is
A. `O_(2)`
B. He
C. `N_(2)`
D. `CO_(2)`
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Correct Answer - B
Given that , `T=27^(@)C=300K,v_("rms")=1365ms^(-1)`
we Know that , `"rms"=sqrt((3RT)/(M))`
`"or " v^(2)"rms"=(3RT)/(M)`
`"or "M+(3RT)/(v^(2)"rms")=(3xx8.31xx300)/(1365xx1365)kg`
`=(3xx8.31xx300)/(1365xx1365)xx1000g~~4g`
The molecular weight of helium is 4 .
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