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:

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