Question

Suppose that the average speed (vrms) of carbon dioxide molecules (molar mass 44.0 g/mol) in a flame is found to be 2.67 105 m/s. What temperature does this represent?

Answer 1

Answer:

$T = 1.26 \times 10^8 K$

Explanation:

As we know that rms speed of ideal gas is given by the formula

$v_{rms} = \sqrt{\frac{3RT}{M}}$

here we know that

$v_{rms} = 2.67 \times 10^5 m/s$

molecular mass of gas is given as

$M = 44 g/mol = 0.044 kg/mol$

now from above formula we have

$2.67\times 10^5 = \sqrt{\frac{3(8.31)T}{0.044}}$

now we have

$T = 1.26 \times 10^8 K$