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Learning Objectives
By the end of this section, you will be able to:
- Describe the distribution of molecular speeds in an idealgas
- Find the average and most probable molecular speeds in an idealgas
Particles in an ideal gas all travel at relatively high speeds,but they do not travel at the same speed. The rms speed is one kindof average, but many particles move faster and many move slower.The actual distribution of speeds has several interestingimplications for other areas of physics, as we will see in laterchapters.
The Maxwell-Boltzmann Distribution
The motion of molecules in a gas is random in magnitude anddirection for individual molecules, but a gas of many molecules hasa predictable distribution of molecular speeds. This predictabledistribution of molecular speeds is known as theMaxwell-Boltzmann distribution, afterits originators, who calculated it based on kinetic theory, and ithas since been confirmed experimentally (Figure\(\PageIndex{1}\)).
To understand this figure, we must define a distributionfunction of molecular speeds, since with a finite number ofmolecules, the probability that a molecule will have exactly agiven speed is 0.
We define the distribution function \(f(v)\) by saying that theexpected number \(N(v_1, v_2)\) of particles with speeds between\(v_1\) and \(v_2\) is given by
\[N(v_1,v_2) = N\int_{v_1}^{v_2} f(v)dv. \nonumber\]
[Since N is dimensionless, the unit off(v) is seconds per meter.] We can write thisequation conveniently in differential form: \[dN = Nf(v)dv.\]
In this form, we can understand the equation as saying that thenumber of molecules with speeds between v and \(v + dv\)is the total number of molecules in the sample timesf(v) times dv. That is, the probabilitythat a molecule’s speed is between v and \(v + dv\) isf(v)dv.
We can now quote Maxwell’s result, although the proof is beyondour scope.
Maxwell-Boltzmann Distribution of Speeds
The distribution function for speeds of particles in an idealgas at temperature \(T\) is
\[f(v) =\dfrac{4}{\sqrt{\pi}}\left(\dfrac{m}{2k_BT}\right)^{3/2}v^2e^{-mv^2/2k_BT}.\]
The factors before the \(v^2\) are a normalization constant;they make sure that \(N(0,\infty ) = N\) by making sure that\(\int_0^{|infty} f(v)dv = 1.\) Let’s focus on the dependence onv. The factor of \(v^2\) means that \(f(0) = 0\) and forsmall v, the curve looks like a parabola. The factor of\(e^{-m_0v^2/2k_BT}\) means that \(\lim_{v\rightarrow \infty} f(v)= 0\) and the graph has an exponential tail, which indicates that afew molecules may move at several times the rms speed. Theinteraction of these factors gives the function the single-peakedshape shown in the figure.
Example \(\PageIndex{1}\): Calculating theRatio of Numbers of Molecules Near Given Speeds
In a sample of nitrogen \(N_2\) with a molar mass of 28.0 g/mol)at a temperature of \(273^oC\) find the ratio of the number ofmolecules with a speed very close to 300 m/s to the number with aspeed very close to 100 m/s.
Strategy
Since we’re looking at a small range, we can approximate thenumber of molecules near 100 m/s as \(dN_{100} = f(100 \, m/s)dv\).Then the ratio we want is
\[\dfrac{dN_{300}}{dN_{100}} = \dfrac{f(300 \, m/s)dv}{f(100 \,m/s)dv} = \dfrac{f(300 \, m/s)}{f(100 \, m/s)}. \nonumber\]
All we have to do is take the ratio of the two fvalues.
Solution
- Identify the knowns and convert to SI units if necessary. \[T =300 \, K, \, k_B = 1.38 \times 10^{-23} J/K\] \[M = 0.0280 \,kg/mol \, so \, m = 4.65 \times 10^{-26} \, kg\]
- Substitute the values and solve. \[\begin{align*} \dfrac{f(300\, m/s)}{f(100 \, m/s)} &=\dfrac{\frac{4}{\sqrt{\pi}}\left(\frac{m}{2k_BT}\right)^{3/2} (300\, m/s)^2 exp[-m(300 \,m/s)^2/2k_BT]}{\frac{4}{\sqrt{\pi}}\left(\frac{m}{2k_BT}\right)^{3/2}(100 \, m/s)^2 exp[-m(100 \, m/s)^2 /2k_BT]} \\[4pt] &=\dfrac{(300 \, m/s)^2 exp [-(4.65 \times 10^{-26} \, kg)(300 \,m/s)^2 /2(1.38 \times 10^{-23} J/K) (300 \, K)]}{(100 \, m/s)^2 exp[-(4.65 \times 10^{-26} \, kg)(100 \, m/s)^2 /2(1.38 \times10^{-23} J/K) (300 \, K)]} \\[4pt] &= 3^2 exp\left[-\dfrac{(4.65 \times 10^{-26} kg)[(300 \, m/s)^2 - (100 \,m/s)^2]}{2(1.38 \times 10^{-23} \, J/K)(300 \, K)} \right] = 5.74\end{align*}\]
Figure \(\PageIndex{2}\) shows that the curve is shifted tohigher speeds at higher temperatures, with a broader range ofspeeds.
With only a relatively small number of molecules, thedistribution of speeds fluctuates around the Maxwell-Boltzmanndistribution. However, you can view this simulation to see theessential features that more massive molecules move slower and havea narrower distribution. Use the set-up “2 Gases, Random Speeds”.Note the display at the bottom comparing histograms of the speeddistributions with the theoretical curves.
We can use a probability distribution to calculate averagevalues by multiplying the distribution function by the quantity tobe averaged and integrating the product over all possible speeds.(This is analogous to calculating averages of discretedistributions, where you multiply each value by the number of timesit occurs, add the results, and divide by the number of values. Theintegral is analogous to the first two steps, and the normalizationis analogous to dividing by the number of values.) Thus the averagevelocity is
\[\overline{v} = \int_0^{\infty} vf(v)dv = \sqrt{\dfrac{8}{\pi}\dfrac{k_BT}{m}} = \sqrt{\dfrac{8}{\pi} \dfrac{RT}{M}}.\]
Similarly,
\[v_{rms} = \sqrt{\overline{v}^2} = \sqrt{\int_0^{\infty} v^2f(v)dv} = \sqrt{\dfrac{3k_BT}{m}} = \sqrt{\dfrac{3RT}{M}}\]
as in Pressure, Temperature, and RMS Speed. The mostprobable speed, also called the peakspeed \(v_p\) is the speed at the peak of thevelocity distribution. (In statistics it would be called the mode.)It is less than the rms speed \(v_{rms}\). The most probable speedcan be calculated by the more familiar method of setting thederivative of the distribution function, with respect tov, equal to 0. The result is
\[v_p = \sqrt{\dfrac{2k_BT}{m}} = \sqrt{\dfrac{2RT}{M}},\]
which is less than \(v_{rms}\). In fact, the rms speed isgreater than both the most probable speed and the averagespeed.
The peak speed provides a sometimes more convenient way to writethe Maxwell-Boltzmann distribution function:
\[f(v) = \dfrac{4v^2}{\sqrt{\pi}v_p^3}e^{-v^2/v_p^2}\]
In the factor \(e^{-mv^2/2k_BT}\), it is easy to recognize thetranslational kinetic energy. Thus, that expression is equal to\(e^{-K/k_BT}\). The distribution f(v) can betransformed into a kinetic energy distribution by requiring that\(f(K)dK = f(v)dv\). Boltzmann showed that the resulting formula ismuch more generally applicable if we replace the kinetic energy oftranslation with the total mechanical energy E.Boltzmann’s result is
\[f(E) = \dfrac{2}{\sqrt{\pi}}(k_BT)^{-3/2}\sqrt{E}e^{-E/k_BT} =\dfrac{2}{\sqrt{\pi}(k_BT)^{3/2}}\dfrac{\sqrt{E}}{e^{E/k_BT}}.\]
The first part of this equation, with the negative exponential,is the usual way to write it. We give the second part only toremark that \(e^{E/k_BT}\) in the denominator is ubiquitous inquantum as well as classical statistical mechanics.
Problem-Solving Strategy: SpeedDistribution
- Step 1. Examine the situation to determinethat it relates to the distribution of molecular speeds.
- Step 2. Make a list of what quantities aregiven or can be inferred from the problem as stated (identify theknown quantities).
- Step 3. Identify exactly what needs to bedetermined in the problem (identify the unknown quantities). Awritten list is useful.
- Step 4. Convert known values into proper SIunits (K for temperature, Pa for pressure, \(m^3\) for volume,molecules for N, and moles for n). In many cases,though, using R and the molar mass will be more convenientthan using \(k_B\) and the molecular mass.
- Step 5. Determine whether you need thedistribution function for velocity or the one for energy, andwhether you are using a formula for one of the characteristicspeeds (average, most probably, or rms), finding a ratio of valuesof the distribution function, or approximating an integral.
- Step 6. Solve the appropriate equation for theideal gas law for the quantity to be determined (the unknownquantity). Note that if you are taking a ratio of values of thedistribution function, the normalization factors divide out. Or ifapproximating an integral, use the method asked for in theproblem.
- Step 7. Substitute the known quantities, alongwith their units, into the appropriate equation and obtainnumerical solutions complete with units.
We can now gain a qualitative understanding of a puzzle aboutthe composition of Earth’s atmosphere. Hydrogen is by far the mostcommon element in the universe, and helium is by far thesecond-most common. Moreover, helium is constantly produced onEarth by radioactive decay. Why are those elements so rare in ouratmosphere? The answer is that gas molecules that reach speedsabove Earth’s escape velocity, about 11 km/s, can escape from theatmosphere into space. Because of the lower mass of hydrogen andhelium molecules, they move at higher speeds than other gasmolecules, such as nitrogen and oxygen. Only a few exceed escapevelocity, but far fewer heavier molecules do. Thus, over thebillions of years that Earth has existed, far more hydrogen andhelium molecules have escaped from the atmosphere than othermolecules, and hardly any of either is now present.
We can also now take another look at evaporative cooling, whichwe discussed in the chapter on temperature and heat. Liquids, likegases, have a distribution of molecular energies. Thehighest-energy molecules are those that can escape from theintermolecular attractions of the liquid. Thus, when some liquidevaporates, the molecules left behind have a lower average energy,and the liquid has a lower temperature.
