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The average kinetic energy of gas molecule at `27^(@)C` is `6.21xx10^(-21)` J. Its average kinetic energy at `127^(@)C` will beA. `12.2xx10^(-21)J`B. `8.28xx10^(-21)J`C. `10.35xx10^(-21)J`D. `11.35xx10^(-21)J` |
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Answer» Correct Answer - B as, KE `prop`T `rArr E_(127)/E_(27)`=((127+273))/((27+273))` `E_(127) =6.21xx120^(-21)xx(400)/(300)` `=.28xx10^(-21) J` |
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