Question

One mole of hydrogen gas is moving at a temperature of 26.7 C. What is its total translational kinetic energy?

Answer 1

`"3740 J"`, or `"3.74 kJ"`.

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At most temperatures, translational energy levels are extremely dense, so the high temperature limit applies essentially all the time for translational motion.

Thus, we can apply the equipartition theorem for the average kinetic energy:

`K_(avg) = N/2 nRT`,

where `N` is the number of degrees of freedom for motion, `n` is mols of gas, `R` is the universal gas constant, and `T` is temperature in `"K"`.

Since `"H"_2` is translating in three dimensions (as we all should), `N = 3`. We also know that `n = "1 mol"` and `T = 26.7 + "273.15 K"`.

So, its total average translational kinetic energy (of an ensemble of `"H"_2` molecules) is given by:

`K_(avg,trans) = 3/2 nRT`

`= 3/2 xx cancel"1 mol" xx "8.314472 J""/"cancel"mol"cdotcancel"K" xx 299.85 cancel"K"`

`=` `3739.64 cdots` `"J"`

To three sig figs, it would be `"3740 J"`, or `color(blue)("3.74 kJ")`.