Question

Suppose that the average velocity (v rms) of carbon dioxide molecules (molecular mass is equal to 44 g/mole) in a flame is found to be 1050 m/s. What temperature does this represent?

Answer 1

Given:

The molecular mass of the molecule is

`M=44\text{ g/mole}`

The speed of the molecule is

`v=1050\text{ m/s}`

Required: The temperature of the molecule is

Step-by-step explanation:

we know that root mean square velocity of the molecule is given as

`v=\sqrt[2]{(3kN_AT)/(M)}`

Here, k is Boltzmann's constant and m is the mass of the molecule in grams and T is the temperature and

`N_A=6.023*10^(23)\text{ mole}^(-1)`
`k=1.38*10^(-23)\text{ J/K}`

we have to find T then square the above relation and simply for T

`\begin{gathered} v=\sqrt[2]{(3kN_AT)/(M)} \\ v^2=(3kN_AT)/(M) \\ T=(v^2M)/(3kN_A) \end{gathered}`

now change the molecular mass into standard units

`\begin{gathered} M=44\text{ g/mole}*\frac{1\text{ kg}}{10^3\text{ g}} \\ M=0.044\text{ kg/mole} \end{gathered}`

now plugging all the values in the above relation we get

`\begin{gathered} T=(v^(2)M)/(3kN_(A)) \\ T=\frac{(1050\text{ m/s})^2*0.044\text{ kg/mole }}{3*1.38*10^(-23)\text{ J/K}*6.023*10^(23)\text{ mole}^(-1)} \\ T=(48510)/(24.93522) \\ T=1945.44\text{ K} \end{gathered}`

Thus, the Temperature of the molecule is

`1945.44\text{ K}`