Respuesta :
Given:
The molecular mass of the molecule is
[tex]M=44\text{ g/mole}[/tex]The speed of the molecule is
[tex]v=1050\text{ m/s}[/tex]Required: The temperature of the molecule is
Explanation:
we know that root mean square velocity of the molecule is given as
[tex]v=\sqrt[2]{\frac{3kN_AT}{M}}[/tex]Here, k is Boltzmann's constant and m is the mass of the molecule in grams and T is the temperature and
[tex]N_A=6.023\times10^{23}\text{ mole}^{-1}[/tex][tex]k=1.38\times10^{-23}\text{ J/K}[/tex]we have to find T then square the above relation and simply for T
[tex]\begin{gathered} v=\sqrt[2]{\frac{3kN_AT}{M}} \\ v^2=\frac{3kN_AT}{M} \\ T=\frac{v^2M}{3kN_A} \end{gathered}[/tex]now change the molecular mass into standard units
[tex]\begin{gathered} M=44\text{ g/mole}\times\frac{1\text{ kg}}{10^3\text{ g}} \\ M=0.044\text{ kg/mole} \end{gathered}[/tex]now plugging all the values in the above relation we get
[tex]\begin{gathered} T=\frac{v^{2}M}{3kN_{A}} \\ T=\frac{(1050\text{ m/s})^2\times0.044\text{ kg/mole }}{3\times1.38\times10^{-23}\text{ J/K}\times6.023\times10^{23}\text{ mole}^{-1}} \\ T=\frac{48510}{24.93522} \\ T=1945.44\text{ K} \end{gathered}[/tex]Thus, the Temperature of the molecule is
[tex]1945.44\text{ K}[/tex]