Pressure, Temperature, and RMS Speed — Question 4

Question 4

Matching rms speeds requires temperature to scale with molar mass

Two ideal gases have molar masses $M_A=.00400$ kg/mol and $M_B=.0280$ kg/mol. Gas A is at 100 K. Use $v_{\mathrm{rms}}=\sqrt{3RT/M}$ with $R=8.314$ J/(mol K).

1. Find the temperature at which B has the same rms speed as A, and calculate that common speed.

2. At equal temperature, find the speed ratio $v_{{\mathrm{rms}},A}/v_{{\mathrm{rms}},B}$. Does equal rms speed imply thermal equilibrium?

Question 4 - Solution

Matching rms speeds requires temperature to scale with molar mass

(a) Method and calculation

$$\frac{T_B}{M_B}=\frac{T_A}{M_A}\ \Rightarrow\ T_B=100\frac{.0280}{.00400}=700\,\mathrm K.$$$$v_{\mathrm{rms}}=\sqrt{\frac{3(8.314)(100)}{.00400}}=\sqrt{623550}\approx789.652\,\mathrm{m/s}.$$

(b) Interpretation and further calculation

$$\left.\frac{v_{{\mathrm{rms}},A}}{v_{{\mathrm{rms}},B}}\right|_{T_A=T_B}=\sqrt{M_B/M_A}=\sqrt7\approx2.64575.$$$$v_{{\mathrm{rms}},A}=v_{{\mathrm{rms}},B}\not\Rightarrow T_A=T_B.$$The equal-speed pair at 100 K and 700 K is not in thermal equilibrium.

Physical check

Mass in the molar formula is kg/mol; using g/mol would introduce a factor $\sqrt{1000}$ error.