A gas occupying a volume of `100` litres is at `20^(@)C` under a pressure of `2` bar. What temperature will it have when it is placed in an evacuated chamber of volume `175` litres? The pressure of the gas in the chamber is one-third of its initial pressure.

Explanation

From the The available data :
`V_(1) = 100 L` , `V_(2) = 175 L`
`P_(1) = 2` bar , `P_(2) = 2 xx 1//3 = 2//3` bar
`T_(1) = 20 + 273 = 293 K` , `T_(2) = ?`
According to 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))`
By substituting the values, `T_(2) = ((2//3 "bar") xx (175 L) xx (293 K))/((2 "bar") xx (100 L))`
`= 170.9 K = 170.9 - 273.0 = - 102.1^(@)C`