About 0.014 kg nitrogen is enclosed in a vessel at temperature of `27^(@)C` How much heat has to be transferred to the gas to double the rms speed of its molecules ? `(R = 2 cal//mol K)`

Explanation

Here, `m=14 g = 0.014 kg`,
`T_(1) = 27^(@)C = (27 + 273) = 300K`
`Q=?` as `c prop sqrt(T)` , therefore , to double the rms speed, temperature T has to be made 4 times , i.e.,
`T_(2) = 4 T_(1) = 4 xx 300 = 1200 K`
Rise in temperature `Delta T = T_(2) - T_(1)`
`= 1200 - 300 = 900 K`
Number of moles , `n=(0.014 xx 1000 g)/(28 g) = 0.5`
As the gas is enclosed in a vessel, it is heated at constant volume. For nitrogen, which is diatomic,
`C_(upsilon) = 5/2 R = 5/2 xx 8.31 J//"mole"//K`
`:. Q = nC_(upsilon) (Delta T)`
`Q= 0.5 xx 5/2 xx 8.31 xx 900 = 9360 J`.