Suppose a 375 ml sample of neon gas at `78^(@)C` is cooled to `22^(@)C` at constant pressure. What will be the new volume of neon sample ?

Explanation

Volume of the neon gas `V_(1)=375 ml`
of a temperature `T_(1)=78^(@)C+273`
At a temperature of `T_(2)=22^(@)C+273`
`=295K`
The valumeof neon gas `V_(2) ?`
According to charles law
`(V_(1))/(T_(1))=(V_(2))/(T_(2))`
`V_(2)=(V_(1)T_(2))/(T_(1))`
`V_(2)=(375xx295)/(351)j=315.17 ml`
`:.` Volume of neon gas at a temperature of `22^(@)C=315.17 ml`