2.2 Pressure, temperature, and rms speed  (Page 4/18)

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${v}_{\text{rms}}=\sqrt{\frac{3\phantom{\rule{0.2em}{0ex}}RT}{M}}.$

We digress for a moment to answer a question that may have occurred to you: When we apply the model to atoms instead of theoretical point particles, does rotational kinetic energy change our results? To answer this question, we have to appeal to quantum mechanics. In quantum mechanics, rotational kinetic energy cannot take on just any value; it’s limited to a discrete set of values, and the smallest value is inversely proportional to the rotational inertia. The rotational inertia of an atom is tiny because almost all of its mass is in the nucleus, which typically has a radius less than ${10}^{-14}\phantom{\rule{0.2em}{0ex}}\text{m}$ . Thus the minimum rotational energy of an atom is much more than $\frac{1}{2}\phantom{\rule{0.2em}{0ex}}{k}_{\text{B}}T$ for any attainable temperature, and the energy available is not enough to make an atom rotate. We will return to this point when discussing diatomic and polyatomic gases in the next section.

Calculating kinetic energy and speed of a gas molecule

(a) What is the average kinetic energy of a gas molecule at $20.0\phantom{\rule{0.2em}{0ex}}\text{ºC}$ (room temperature)? (b) Find the rms speed of a nitrogen molecule $\left({\text{N}}_{2}\right)$ at this temperature.

Strategy

(a) The known in the equation for the average kinetic energy is the temperature:

$\stackrel{\text{–}}{K}=\frac{1}{2}\phantom{\rule{0.2em}{0ex}}m\stackrel{\text{–}}{{v}^{2}}=\frac{3}{2}{k}_{\text{B}}T.$

Before substituting values into this equation, we must convert the given temperature into kelvin: $T=\left(20.0+273\right)\phantom{\rule{0.2em}{0ex}}\text{K}=293\phantom{\rule{0.2em}{0ex}}\text{K}\text{.}$ We can find the rms speed of a nitrogen molecule by using the equation

${v}_{\text{rms}}=\sqrt{\stackrel{\text{–}}{{v}^{2}}}=\sqrt{\frac{3{k}_{\text{B}}T}{m}},$

but we must first find the mass of a nitrogen molecule. Obtaining the molar mass of nitrogen ${\text{N}}_{2}$ from the periodic table, we find

$m=\frac{M}{{N}_{A}}=\frac{2\phantom{\rule{0.2em}{0ex}}\left(14.0067\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\phantom{\rule{0.2em}{0ex}}\text{kg/mol}\right)}{6.02\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{23}\phantom{\rule{0.2em}{0ex}}{\text{mol}}^{\text{-1}}}=4.65\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-26}\phantom{\rule{0.2em}{0ex}}\text{kg}.$

Solution

1. The temperature alone is sufficient for us to find the average translational kinetic energy. Substituting the temperature into the translational kinetic energy equation gives
$\stackrel{\text{–}}{K}=\frac{3}{2}\phantom{\rule{0.2em}{0ex}}{k}_{\text{B}}T=\frac{3}{2}\left(1.38\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-23}\phantom{\rule{0.2em}{0ex}}\text{J/K}\right)\left(293\phantom{\rule{0.2em}{0ex}}\text{K}\right)=6.07\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-21}\phantom{\rule{0.2em}{0ex}}\text{J}.$
2. Substituting this mass and the value for ${k}_{\text{B}}$ into the equation for ${v}_{\text{rms}}$ yields
${v}_{\text{rms}}=\sqrt{\frac{3{k}_{\text{B}}T}{m}}=\sqrt{\frac{3\left(1.38\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-23}\phantom{\rule{0.2em}{0ex}}\text{J/K}\right)\left(293\phantom{\rule{0.2em}{0ex}}\text{K}\right)}{4.65\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-26}\phantom{\rule{0.2em}{0ex}}\text{kg}}}=511\phantom{\rule{0.2em}{0ex}}\text{m/s}\text{.}$

Significance

Note that the average kinetic energy of the molecule is independent of the type of molecule. The average translational kinetic energy depends only on absolute temperature. The kinetic energy is very small compared to macroscopic energies, so that we do not feel when an air molecule is hitting our skin. On the other hand, it is much greater than the typical difference in gravitational potential energy when a molecule moves from, say, the top to the bottom of a room, so our neglect of gravitation is justified in typical real-world situations. The rms speed of the nitrogen molecule is surprisingly large. These large molecular velocities do not yield macroscopic movement of air, since the molecules move in all directions with equal likelihood. The mean free path (the distance a molecule moves on average between collisions, discussed a bit later in this section) of molecules in air is very small, so the molecules move rapidly but do not get very far in a second. The high value for rms speed is reflected in the speed of sound, which is about 340 m/s at room temperature. The higher the rms speed of air molecules, the faster sound vibrations can be transferred through the air. The speed of sound increases with temperature and is greater in gases with small molecular masses, such as helium (see [link] ).

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