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

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 ${10}^{\mathrm{-14}}\text{m}$ . Thus the minimum rotational energy of an atom is much more than $\frac{1}{2}{k}_{\text{B}}T$ 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

Strategy

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

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

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
   
   $\stackrel{\text{–}}{K}=\frac{3}{2}{k}_{\text{B}}T=\frac{3}{2}(1.38\times {10}^{\mathrm{-23}}\text{J/K})(293\text{K})=6.07\times {10}^{\mathrm{-21}}\text{J}.$
2. Substituting this mass and the value for $k_{\text{B}}$ into the equation for $v_{\text{rms}}$ yields
   
   $v_{\text{rms}}=\sqrt{\frac{3k_{\text{B}}T}{m}}=\sqrt{\frac{3(1.38\times {10}^{\mathrm{-23}}\text{J/K})(293\text{K})}{4.65\times {10}^{\mathrm{-26}}\text{kg}}}=511\text{m/s}\text{.}$

Significance