A sample of gas occupies volume of `2.74 dm^(3)` at `0.9` bar and `27^(@)C`. What will be the volume at `0.75` bar and `15^(@)C`?

Explanation

Form the available data : `V_(1) = 2.74 dm^(3)` , `V_(2) = ?`
`P_(1) = 0.9` bar , `P_(2) = 0.75` bar
`T_(1) = 27 + 273 = 300 K` , `T_(2) = 15 + 273 = 288 K`
According to gas equation, `(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))` or `V_(2) = (P_(1)V_(1)T_(2))/(T_(1)P_(2))`
By substituting the values,
`V_(2) = ((0.9 "bar") xx (2.74 dm^(3)) xx (288 K))/((300 K) xx (0.75 "bar" )) = 3.16 dm^(3)`