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.