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The temperature at which mean kinetic energy of an ideal gas molecule will be doubled that at `27^(@)C` isA. `54^(@)C`B. `13.5^(@)C`C. `600^(@)K`D. `150^(@)K` |
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Answer» Correct Answer - C `K.E_(2) prop T_(1) and KE_(2) prop T_(2)` `(KE_(2))/(KE_(1))=(T_(2))/(T_(1))` `therefore (2KE_(1))/(KE_(1))xxT_(1)=T_(2)` `2xx300=T_(2)` `therefore T_(2)=600K` |
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