Answer to Question #160211 in Molecular Physics | Thermodynamics for Yolande

Question

A gas is at 200K. If we wish to double the rms speed of the molecules of the gas, to what value must we raise its temperature?

Answer 1

The rms speed of the molecules of the gas can be found as follows:

$v_{rms}=\sqrt{\dfrac{3RT}{M}}.$

Let the initial rms speed of the molecules of the gas will be:

$v_{rms,1}=\sqrt{\dfrac{3RT_1}{M}}.$

Let the final rms speed of the molecules of the gas will be:

$v_{rms,2}=\sqrt{\dfrac{3RT_2}{M}}$

Then, dividing the first expression by the second one and assuming that $v_{rms,2}=2v_{rms,1}$, we get:

$\dfrac{v_{rms,1}}{2v_{rms,1}}=\sqrt{\dfrac{T_1}{T_2}},$ $\dfrac{1}{4}=\dfrac{T_1}{T_2},$ $T_2=4T_1=4\cdot200\ K=800\ K.$