Q45PE - College Physics (Urone)

Question

Question: Suppose that the average velocity (v_rms) of carbon dioxide molecules (molecular mass is equal to 44.0 g/mol) in a flame is found to be 1.05×10⁵ m/s. What temperature does this represent?

Step-by-Step Solution

Answer

Answer

The temperature at which the rms velocity of carbon dioxide molecules is 1.05×10⁵ m/s is 1.94×10⁷ K.

Step 1: Introduction

We apply the formula for rms velocity to find the temperature of the carbon dioxide molecules in the flame.

Step 2: Given parameters and formula for rms velocity

rms velocity =105000m/s

Molar mass =44g/mol

The formula for rms velocity${\text{v}}_{\text{rms}}\text{ = }\sqrt{\frac{\text{3kT}}{\text{m}}}$

Here, \({{\rm{v}}_{{\rm{rms}}}}\) rms velocity, \({\rm{m,k}}\) are the mass of an atom/molecule and Boltzmann constant, respectively, and \({\rm{T}}\)  temperature.

Step 3: calculate the mass of a molecule and temperature

$$\begin{gathered} \mathrm{Massofacarbondioxidemolecule}=\frac{44\times {10}^{-3}}{6.023\times {10}^{23}}\mathrm{kg} \\ =7.305\times {10}^{-26}\mathrm{kg} \end{gathered}$$

$$\begin{gathered} {}_{\text{rms}}=\sqrt{\frac{\text{3kT}}{m}} \\ {{\text{v}}_{\text{rms}}}^{\text{2}}=\frac{\text{3kT}}{m} \\ \text{T}=\frac{\text{m}{{\text{v}}_{\text{rms}}}^{\text{2}}}{\text{3k}} \\ T=\frac{ (7.305\times {10}^{-26}Kg)\times (105000 m/s^2)}{3\times (1.38\times {10}^{-23}J/k)} \\ \text{T}=\text{1.94}\times {\text{10}}^{\text{7}} \text{K} \end{gathered}$$

The temperature at which the rms velocity of carbon dioxide molecules is 1.05×10⁵ m/s is