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The ratio between enrgies of `16 g` of `O_(2)` and `28 g` of `N_(2)` respectively at `300 K` will beA. `1:1`B. `1:2`C. `2:1`D. `4:7` |
Answer» Correct Answer - B Number of moles of `O_(2)= (16)/(32) =(1)/(2)` Number of moles of `N_(2)=(28)/(28)=1` Ratio between number of moles `(1)/(2):1` `KE` will be in the ratio `1:2` |
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