The volume of a certain quantity of oxygen is `200 cm^(3)` at `15^(@)C` and `0.98` bar pressure of Hg. What should be the temperature so that the volume becomes `180 cm^(3)` under a pressure of `1 "bar"` of Hg?

Answer

`-8.5^(@)C`

Explanation

`V_(200) cm^(3) , V_(2) = 180 cm^(3)`
`P_(1) = 0.098 "bar" , P_(2) = 1.0 "bar"`
`T_(1) = 15 + 273 = 288 K , T_(2)= ?`
According to ideal gas equation,
`(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))` or `T_(2) = (P_(2)V_(2)T_(1))/(P_(1)V_(1)) = ((1 "bar") xx (180 cm^(3)) xx (288 K))/((0.98 "bar") xx (200 cm^(3)) = 264.5 = -2.5^(@)C`