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

Jul 29, 2017

Whatever quadruple the temperature in $\text{K}$ is. Would this apply to RMS speed and average speed as well? Why?

The most probable speed is given by

${v}_{m p} = \sqrt{\frac{2 {k}_{B} T}{m}}$,

where:

• ${k}_{B}$ is the Boltzmann constant in $\text{J/K}$.
• $T$ is temperature in $\text{K}$.
• $m$ is the single-particle mass in $\text{kg}$.

If you want to double the most probable speed as that at ${127}^{\circ} \text{C}$, or $\text{400.15 K}$, you want...

${v}_{m p} ' = 2 {v}_{m p} = 2 \sqrt{\frac{2 {k}_{B} T}{m}}$

$= \sqrt{\frac{2 {k}_{B} \cdot \textcolor{red}{4 T}}{m}}$

i.e. the temperature in $\text{K}$ should quadruple to $\textcolor{b l u e}{\text{1600.60 K}}$.

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

$v \propto \sqrt{T}$

We have...

$2 v \propto \sqrt{4 T}$

$\implies$ temperature in $\text{K}$ must have quadrupled.