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What is the increase in volume, when the temperature of `600 mL` of air increases from `27^(@)C` to `47^(@)C` under constant pressure?A. `80 mL`B. `40 mL`C. `640 mL`D. `500 mL`

Answer» Correct Answer - B
The increase in volume of a sample ………….
`(V_(1))/(T_(1)) = (V_(2))/(T_(2))`
so `V_(2) = (T_(2))/(T_(1)). V_(1) = (320)/(300)xx600 mL = 640 mL`
so increment `= (640-600)mL = 40 mL`


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