Question

Suppose that the typical speed (vrms) of carbon dioxide molecules (molar mass is 44.0 g/mol) in a flame is found to be 1350 m/s. What temperature does this indicate?

a) 500 K
b) 1000 K
c) 1500 K
d) 2000 K

Answer 1

Using the root-mean-square speed equation and the given data, the temperature corresponding to a Vrms of 1350 m/s for carbon dioxide molecules is approximately 2000 K, which is option d.

Step-by-step explanation:

To calculate the temperature that corresponds to a given root-mean-square speed (Vrms) of carbon dioxide molecules in a flame, one can use the formula that connects the speed with temperature and molar mass:

Vrms = sqrt((3kT)/M)

Here, Vrms is the root-mean-square speed, k is the Boltzmann constant (1.38 × 10-23 J/K), T is the temperature in kelvins, and M is the molar mass of the gas.

First, it is necessary to convert the molar mass from grams per mole to kilograms per mole by dividing by 1000 since the SI unit for mass in the formula is kilograms. For carbon dioxide (CO2) with a molar mass of 44.0 g/mol, this gives a mass of 0.044 kg/mol.

Using the given Vrms of 1350 m/s and the formula, we can solve for temperature T:

T = (Vrms2 × M) / (3k)

T = (13502 × 0.044) / (3 × 1.38 × 10-23)

T = ≈ 2000 K

Therefore, the temperature corresponding to the typical speed of carbon dioxide molecules in a flame being 1350 m/s is approximately 2000 K, which corresponds to option d.