When the temperature of a gas increases from `0^(@)C` the valume of the gas increases by a factor of `1.25`what is the final temperature ?

Explanation

Initial temperature `T_(1)` of the gas = `0^(@)C =0+273`
`=273^(@)C`
Let the Initial volumeof the gas `V_(1) =x ml`
The final volume `V_(2)` of the gas `=1.25xx x ml`
Accoridng to Charle's law
`(V_(1))/(T_(1))=(V_(2))/(T_(2))`
`(x)/(273)=(1.25xx x)/(T_(2))`
`T_(2)=(1.25xx cancel x xx 273)/(cancel x)`
`=341.25 K`
The final temperature `=341.25 K`