(2) 25%
Let V and T denote initial volume and temperature of gas. The corresponding quantities finally are V1 , T1 . From Charle’s law V α T (P remaining constant) therefore

V1 would have been the volume of gas if there was leakage in cylinder. However the final volume V´ = V - 0.1V = 0.9V.
∴ Percentage of volume leaked out = \(\frac{(1.2-0.9)V}{1.2V}\times100= 25\%\)