Certain amount of gas occupies a volume of `400 mL` a `17^(@)C`. To What temperature should it is beated so that the volume is reduced to half?

Answer

`-128^(@)C`

Explanation

`V_(1) = 400 mL , V_(2) = ?`
`T_(1)= 17 + 273 = 290 K, T_(2) = ?`
According to Charles Law.
`(V_(1))/(T_(2)) = (V_(2))/(T_(2))` or `T_(2) = (V_(2) xx T_(1))/(V_(1)) = ((200 mL) xx (290 K))/((400 mL)) = 145 K = - 128^(@)C`