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

Question #160211
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?
1
Expert's answer
2021-02-25T15:04:15-0500

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


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

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


vrms,1=3RT1M.v_{rms,1}=\sqrt{\dfrac{3RT_1}{M}}.

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


vrms,2=3RT2Mv_{rms,2}=\sqrt{\dfrac{3RT_2}{M}}

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


vrms,12vrms,1=T1T2,\dfrac{v_{rms,1}}{2v_{rms,1}}=\sqrt{\dfrac{T_1}{T_2}},14=T1T2,\dfrac{1}{4}=\dfrac{T_1}{T_2},T2=4T1=4200 K=800 K.T_2=4T_1=4\cdot200\ K=800\ K.

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