Suppose that the root‑mean‑square velocity vrms of carbon dioxide molecules (molecular mass is equal to 44.0 g/mol ) in a flame is found to be 1310 m/s. What temperature T does this represent? The Boltzmann constant is k=1.38×10−23 J/K and Avogadro's number is ????A=6.022×1023 mol−1.

Respuesta :

Answer : The value of temperature is, 3028.7 K

Explanation :

The formula used for root mean square speed is:

[tex]\nu_{rms}=\sqrt{\frac{3kN_AT}{M}}[/tex]

where,

[tex]\nu_{rms}[/tex] = root mean square speed  = 1310 m/s

k = Boltzmann’s constant = [tex]1.38\times 10^{-23}J/K[/tex]

T = temperature = ?

M = molar mass = 44.0 g/mole  = 0.044 kg/mol

[tex]N_A[/tex] = Avogadro’s number = [tex]6.022\times 10^{23}mol^{-1}[/tex]

Now put all the given values in the above root mean square speed formula, we get:

[tex]1310m/s=\sqrt{\frac{3\times (1.38\times 10^{-23}J/K)\times (6.022\times 10^{23}mol^{-1})\times (T)}{0.044kg/mol}}[/tex]

[tex]T=3028.7K[/tex]

Therefore, the value of temperature is, 3028.7 K

The root mean square velocity is directly proportional to the temperature's square root and is inversely proportional to the square root of the molar mass.

The value of the temperature of the molecules is 3028.7 K.

Given:

Root mean square speed  = 1310 m/s

Boltzmann’s constant = 1.38×10⁻²³

Molar mass = 0.044 kg/mol

Avogadro’s number = 6.022×10²³ mol⁻¹

Temperature =?

The formula for root mean square is:

V[tex]_\text{rms}[/tex] = [tex]\sqrt{\dfrac{3\text {k N}_\text A \text T}{\text M}[/tex]

[tex]1310 = \sqrt{\dfrac{3 \times 1.38 \times 10 ^{-23} \times 6.022 \times 10 ^{23} \times \text T}{0.44}[/tex]

T = 3028.7 K

Therefore, the temperature for the reaction is 3028.7 K.

To know more about root mean square velocity, refer to the following link:

https://brainly.com/question/2020688