Compute the temperature at which the r.m.s speed of nitrogen molecules is `832m//s`. [Universal gas constant, `R=8320J//k` mole `K`, molecular weight of nitrogen `=28`.]

Explanation

Given :
`v_(r.m.s)=832ms^(-1)`
`R=8320 J//K mol K`
`M=28`
`v_(r.m.s.)=sqrt((3RT)/(M))`
`832=sqrt((3xx8320xxt)/(28))`
`T=(832xx832xx28)/(3xx8320)`
`T=776.53K`
or `T=503.53^(@)C`.