# 16.2 ROOT MEAN SQUARE SPEED OF AN IDEAL GAS The tiny molecules in a gas are all zipping around at different speeds, some are fast, and some are slow. So, how can we talk about their "average" speed in a useful way? This section introduces the **root mean square (rms) speed**, a special kind of average that gives us a true picture of the energy in a gas. We'll explore what this speed means, how it's calculated, and see how it is directly linked to the gas's temperature and the mass of its molecules. ### Kinetic Theory of Gases Explains that molecules in a gas move at different velocities. ### Mean Square Speed ($\langle v^{2} \rangle$) The average of the squares of the speeds of all gas molecules. $$ \langle v^{2} \rangle = \frac{v_{1}^{2}+v_{2}^{2}+v_{3}^{2}+\ldots .+v_{N}^{2}}{N} $$ ### Root Mean Square (rms) Speed ($v_{r.m.s}$) The square root of the mean square speed of gas molecules. It's a measure of the typical speed of molecules. $$ v_{r.m.s} = \sqrt{\frac{v_{1}^{2}+v_{2}^{2}+v_{3}^{2}+\ldots .+v_{N}^{2}}{N}} \quad \text{(Eq. 16.17)} $$ ### Average Speed ($v_{av}$) The arithmetic mean of the speeds of the molecules of a gas at a given temperature. $$ v_{av} = \frac{v_{1}+v_{2}+v_{3}+\ldots .+v_{N}}{N} \quad \text{(Eq. 16.18)} $$ ### Derivation of $v_{r.m.s}$ from Pressure Equation * Starting from the pressure equation derived in , $P = \frac{1}{3} \frac{m N}{V} \langle v^{2} \rangle$, which can be rearranged to $PV = \frac{m N}{3} \langle v^{2} \rangle$ (Eq. 16.19). * Comparing with the ideal gas law $PV = NkT$ (Eq. 16.20), we get: $$ \frac{mN}{3} \langle v^{2} \rangle = NkT \quad \implies \quad \langle v^{2} \rangle = \frac{3 kT}{m} \quad \text{(Eq. 16.21)} $$ * Taking the square root gives the rms speed: $$ v_{r.m.s} = \sqrt{\frac{3 k T}{m}} \quad \text{(Eq. 16.22)} $$ where $k$ is the Boltzmann constant, $T$ is the absolute temperature, and $m$ is the mass of one molecule. ### Relationship of $v_{r.m.s}$ with Temperature * $v_{r.m.s} \propto \sqrt{T}$ * The root mean square speed is directly proportional to the square root of the absolute temperature. ### $v_{r.m.s}$ in terms of Molar Mass ($M$) * Substituting $k = \frac{R}{N_A}$ (where $R$ is the ideal gas constant and $N_A$ is Avogadro's number) and $mN_A = M$ (molar mass) into Eq. (16.22): $$ v_{r.m.s} = \sqrt{\frac{3 R T}{M}} \quad \text{(Eq. 16.23)} $$ ### Relationship of $v_{r.m.s}$ with Molar Mass * At constant temperature, $v_{r.m.s} \propto \frac{1}{\sqrt{M}}$ * The root mean square speed is inversely proportional to the square root of the molar mass. Lighter gases have higher rms speeds. ### $v_{r.m.s}$ in terms of Pressure ($P$) and Density ($\rho$) * From the pressure equation $P = \frac{1}{3} \rho \langle v^{2} \rangle$: $$ \langle v^{2} \rangle = \frac{3 P}{\rho} $$ * Taking the square root gives: $$ v_{r.m.s} = \sqrt{\frac{3 P}{\rho}} \quad \text{(Eq. 16.24)} $$ ### Summary of $v_{r.m.s}$ Formulas (Eq. 16.25) $$ v_{r.m.s} = \sqrt{\frac{3 k T}{m}} = \sqrt{\frac{3 R T}{M}} = \sqrt{\frac{3 P}{\rho}} $$ ### Boltzmann Constant ($k$) values * $k = 1.38 \times 10^{-16} \text{ erg K}^{-1}$ * $k = 1.38 \times 10^{-23} \text{ J K}^{-1}$ * $k = 3.3 \times 10^{-24} \text{ cal K}^{-1}$ ## **Possible Questions/Answer** - **Q:** What is the root mean square speed of oxygen molecules in the air at standard temperature and pressure (STP)? **A:** Given $T = 273 \text{ K}$, $P = 1.01 \times 10^5 \text{ Pa}$. Molar mass of oxygen $M = 32 \text{ g/mol} = 32 \times 10^{-3} \text{ kg/mol}$. Mass of one oxygen molecule $m = \frac{M}{N_A} = \frac{32 \times 10^{-3} \text{ kg}}{6.022 \times 10^{23}} = 5.31 \times 10^{-26} \text{ kg}$. Using $v_{r.m.s} = \sqrt{\frac{3kT}{m}}$: $v_{r.m.s} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \text{ J/K} \times 273 \text{ K}}{5.31 \times 10^{-26} \text{ kg}}} = 461.4 \text{ m/s}$. - **Q:** Four molecules have speeds of $2 \text{ km s}^{-1}, 4 \text{ km s}^{-1}, 6 \text{ km s}^{-1}$, and $8 \text{ km s}^{-1}$. Calculate their average speed. **A:** $v_{av} = \frac{v_1+v_2+v_3+v_4}{N} = \frac{2+4+6+8}{4} = \frac{20}{4} = 5 \text{ km/s}$. - **Q:** For the same four molecules (speeds $2, 4, 6, 8 \text{ km s}^{-1}$), calculate their root mean square speed. **A:** $v_{r.m.s} = \sqrt{\frac{v_1^2+v_2^2+v_3^2+v_4^2}{N}} = \sqrt{\frac{(2)^2+(4)^2+(6)^2+(8)^2}{4}} = \sqrt{\frac{4+16+36+64}{4}} = \sqrt{\frac{120}{4}} = \sqrt{30} \approx 5.48 \text{ km/s}$. - **Q:** How does the root mean square speed of a gas relate to its molar mass at constant temperature? **A:** At constant temperature, $v_{r.m.s} \propto \frac{1}{\sqrt{M}}$. The root mean square speed is inversely proportional to the square root of the molar mass of the gas. Lighter gases move faster. | Formula Name | Expression | | :----------------------- | :------------------------------------------------------------------------------- | | Mean Square Speed | $\langle v^{2} \rangle = \frac{\sum v_{i}^{2}}{N}$ | | Root Mean Square Speed | $v_{r.m.s} = \sqrt{\frac{\sum v_{i}^{2}}{N}}$ | | Average Speed | $v_{av} = \frac{\sum v_{i}}{N}$ | | $v_{r.m.s}$ (k,T,m) | $v_{r.m.s} = \sqrt{\frac{3 k T}{m}}$ | | $v_{r.m.s}$ (R,T,M) | $v_{r.m.s} = \sqrt{\frac{3 R T}{M}}$ | | $v_{r.m.s}$ (P,$\rho$) | $v_{r.m.s} = \sqrt{\frac{3 P}{\rho}}$ | | Boltzmann Constant ($k$) | $1.38 \times 10^{-23} \text{ J K}^{-1}$ (SI) |