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The density of a gas at `27^(@)C` and `1 atm` is `d`. Pressure remaining constant, at which of the following temperture will its density become `0.75 d`?A. `20^(@)C`B. `30^(@)C`C. `400 K`D. `300 K` |
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Answer» `(d)_(1)=d` `(d)_(2)=0.75 d` `P_(1)= 1 atm`, `T_(1)=27^(@)C=300 K` `P_(2)=1 atm`, `T_(2)=?` `((d_(1))/(d_(2)))=(P_(1)xxT_(2))/(T_(1)xxP_(2))` or `(T_(2))/(T_(1))` `:. T_(2)=(d_(1))/(d_(2))xxT_(1)=(d)/(0.75d)xx300=400K` |
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