# Question Video: Calculating the Standard Deviation of a Set of Measurements

The root-mean-square speed of carbon dioxide molecules in a flame is found to be 1350 m/s. Find the temperature of the flame. Use a value of 44.0 g/mol for the molar mass of carbon dioxide molecules.

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### Video Transcript

The root-mean-square speed of carbon dioxide molecules in a flame is found to be 1350 meters per second. Find the temperature of the flame. Use a value of 44.0 grams per mole for the molar mass of carbon dioxide molecules.

Inside this flame, we can imagine the carbon dioxide molecules moving around with a distribution of speeds. And we’re told that the square root of the average of the square of those speeds is 1350 meters per second. We want to solve for the temperature of the flame. And we’ll see that the root-mean-square speed of the molecules in the flame help tell us what that is.

We can rely on the mathematical relationship that says 𝑣 rms for a gas is equal to the square root of three times the gas constant 𝑅 times the temperature of the molecules in the gas all divided by the molar mass of those molecules. If we rearrange this equation to solve for 𝑇, we find it’s equal to the root-mean-square speed of our molecules squared multiplied by the molar mass all over three 𝑅.

We’re told that the molar mass of these carbon dioxide molecules is 44.0 grams per mole. And we can look up the value for the gas constant 𝑅 and find that it’s 8.314 joules per mole Kelvin. When we plug in these values for 𝑀, 𝑅, and 𝑣 sub rms, we’re careful to convert our molar mass 𝑀 into units of kilograms per mole.

Calculating this value to three significant figures, we find it’s 3.22 times 10 to the third Kelvin. That’s the flame temperature that corresponds to this rms speed of carbon dioxide molecules.