What new temperature will a gas have if its speed has to become doubled compared to #127^@ "C"#?

1 Answer
Jul 29, 2017

Whatever quadruple the temperature in #"K"# is. Would this apply to RMS speed and average speed as well? Why?


The most probable speed is given by

#v_(mp) = sqrt((2k_BT)/m)#,

where:

  • #k_B# is the Boltzmann constant in #"J/K"#.
  • #T# is temperature in #"K"#.
  • #m# is the single-particle mass in #"kg"#.

If you want to double the most probable speed as that at #127^@ "C"#, or #"400.15 K"#, you want...

#v_(mp)' = 2v_(mp) = 2sqrt((2k_BT)/m)#

#= sqrt((2k_B cdot color(red)(4T))/m)#

i.e. the temperature in #"K"# should quadruple to #color(blue)("1600.60 K")#.


Or with any other speed that follows from the Maxwell-Boltzmann distribution, i.e. any other speed that has the relationship:

#v prop sqrtT#

We have...

#2v prop sqrt(4T)#

#=># temperature in #"K"# must have quadrupled.