# 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

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

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

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

i.e. the temperature in

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.