An extremely small sample of an ideal diatomic gas (with a molar mass of 28 g/mol)...

An extremely small sample of an ideal diatomic gas (with a molar mass of 28 g/mol) has the following distribution of molecular speeds: 1 molecule moving at 100 m/s, 2 molecules at 200 m/s, 4 at 300 m/s, and 3 at 400 m/s.

What is the rms speed of the distribution? What is the average kinetic energy of translational motion per molecule? What is the temperature of this sample?

Solutions

Expert Solution

The root mean square or rms speed is equal to the square of the sum of the speeds divided by the number of molecules. It is

Rounding off to two significant figures, the rms speed of the distribution is 300 m/s.

The mass of a molecule is

m = 28 g/mol

= 28 g / 6.023 x 10²³

= 0.0046488 x 10^-23 kg

Now, the average kinetic energy of translational motion per molecule is

Substitute 0.0046488 x 10^-23 kg for m and 300 m/s for v_rms in the above equation,

Rounding off to two significant figures, the average kinetic energy of translational motion per molecule is 2.1 x 10^-21 J.

Let T be the temeprature of the sample.

The rms speed of the molecules is

Here, molar mass is M and gas constant is R.

Substitute 300 m/s for v_rms, 28 g/mol for M and 8.31 J/ K mol for R the values in the above equation

Rounding off to two significant figures, the temperature of the sample is 300 K.

genius_generous answered 4 months ago

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