# At what temperature C do the molecules of nitrogen gas have an rms speed of 265 m/s?

Update:

78.8 K = -194.2 C

Relevance

Velocity = sq rt (3RT / M)

N2 = 28 g/mole

Temp = C + 273C = K

R = gas constant = 8.314 J/mol K

265 m/s = sq rt ( 3 * [8.314 kg m^2/ s^2 mol K] * T) / 0.028 kg /mole

265 m/s = sq rt ( 890.79 *T)

square both sides:

(265 m/s)^2 = 890.79 * T

solve for T :

T = (265 m/s)^2 / 890.79

T = 78.8 K

• Anonymous
4 years ago

in accordance to Maxwell Boltzmann kinetic concept of gases; ½ mv² = 33a26aab899028a722fd1f962e68f9a333a26aab... kT as a result, v = ?(14/40 4) 33a26aab899028a722fd1f962e68f9a3 kT33a26aab899028a722fd1f962e68f9am33a26a... and we see that the rms velocity of the gases are inversely proportional to their mass. an hassle-loose share ought to be devised: v'33a26aab899028a722fd1f962e68f9av = ?(14/40 4) 33a26aab899028a722fd1f962e68f9am33a26aab... 4), ....and, v' = 33a26aab899028a722fd1f962e68f9a33a26aab8... = [33a26aab899028a722fd1f962e68f9a33a26aab... = 2933a26aab899028a722fd1f962e68f9a.6 m33a26aab899028a722fd1f962e68f9as