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`152 mL` of a gas at `STP` was taken to `20^(@)C` and `729 mm` pressure. What was the change in volume of the gas? |
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Answer» Given `V_(1)=152 mL` `P_(1)=1 atm=760 mm` `T_(1)=273.15 K` `V_(2)=?` `P_(2)=729 m` `T_(2)=20^(@)C` or `293.15 K` `:. V_(2)=(P_(1)V_(1))/(T_(1))xx(T_(2))/(P_(2))=(760xx152xx293.15)/(273.15xx729)=170 mL` `:. Change in volume=V_(2)-V_(1)=170-152=18 mL` |
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