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A gas at a pressure of `5.0 atm` is heated from `0^(@)C` to `546^(@)C` and simultaneously compressed to one-third of its original volume. Hence final presseure isA. `10.0 atm`B. `30.0 atm`C. `45.0 atm`D. `5.0 atm` |
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Answer» Correct Answer - C `P_(1)= 5.0 atm, T_(1)= 0^(@)C= 273, V_(1)=V` `P_(2)=? T_(2)= 546^(@)C= 819 K, V_(2)= (1)/(3)V` `:. From (P_(1)V_(1))/(P_(2)V_(2))=(T_(1))/(T_(2))` `implies (5xxV)/(P_(2)xx(1)/(3)V)= (273)/(819)` `implies (5xx3)/(P_(2))=(1)/(3)` `P_(2)= 45 atm` |
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