Question

Expressions for the Maxwell speed distribution for molecules in a gas are given in Chapter 19. (a) Show that the most probable energyis given by

${\mathbf{K}}_{\mathbf{P}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{kT}$

Verify this result with the energy distribution curve of Fig. 43-10, for which$\mathit{T}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{\times }\mathbf{10}\mathit{K}$. (b) Show that the most probable speedis given by

${\mathbf{v}}_{\mathbf{P}}\mathbf{=}\sqrt{\frac{\mathbf{2}\mathbf{kT}}{\mathbf{m}}}$

Find its value for protons at$\mathbf{T}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathbf{K}$. (c) Show that the energy corresponding to the most probable speed(which is not the same as the most probable energy) is

${\mathbf{K}}_{\mathbf{V}\mathbf{,}\mathbf{P}}\mathbf{=}\mathbf{kT}$

Locate this quantity on the curve of Fig. 43-10.

Short answer

1. The most probable energy is given by, $K_P=\frac{1}{2}kT$and the value of the energy at the given temperature is$0.65keV$.
2. The most probable speed is given by,$V_P=\sqrt{\frac{2kT}{m}}$and the probable speed of a proton is$5.0\times {10}^5\frac{m}{s}$.
3. The energy corresponding to the most probable speed is$K_{V,P}=kT$and the value of the kinetic energy is$1.3\mathrm{ke}V$.

Step-by-step solution

Write given data

1. The expressions for Maxwell speed distribution for molecules in a gas are given.
2. Energy distribution curve of figure 43-10 is given.
3. The given temperature,$T=1.5x{10}^7\mathrm{K}$

Determine the concept of probable speed and energy

The most probable speed is the speed of the molecules which informs about the speed possessed by the maximum number of molecules of the gas. It is maximum at Maxwell's distribution curve. Similarly, the energy that is possessed by the maximum number of molecules is referred as the most probable energy of the molecules.

The equation for number density per unit energy for particles as per problem 43-34,

$n\left(K\right)=1.13n\frac{K^{1/2}}{{\left(kT\right)}^{{}^{1/2}}{e}^{-K/KT}}$ …… (i)

The most probable speed according to equation 19-35 as follows:

$V_P=\sqrt{\frac{2RT}{M}}$ …...(ii)

The molar mass relation to the given mass of a sample as per equation 19-7,

$M=m{N}_A$ …… (iii)

The kinetic energy of a particle as:

$K=\frac{1}{2}m{v}^2$ …… (iv)

a) Calculate the most probable energy

Rather than use$P\left(v\right)$as it is written in Eq. 19-27, we use the more convenient$n\left(k\right)$expression given in Problem 43-34.

Thus, the most probable energy can be calculated by taking the derivation of the number density per unit energy for particle from equation (i) as follows:

$$\begin{gathered} {\overline{)\frac{dn\left(K\right)}{dK}}}_{k=k_P}=\frac{1.13n}{{\left(kT\right)}^{{\displaystyle \frac{3}{2}}}}\left(\frac{1}{k^{1/2}}-\frac{k^{3/2}}{kT}\right){\overline{)e^{-K/kT}}}_{K=KP} \\ 0=\frac{1.13n}{{\left(kT\right)}^{{\displaystyle 1/2}}}\left(\frac{1}{2{K_P}^{1/2}}-\frac{K_p^2}{kT}\right)e^{-k_p/kT} \\ \frac{1}{2{K_P}^{1/2}}-\frac{2{K_P}^{1/2}}{kT}=0 \\ {k}_p=\frac{1}{2}kT \end{gathered}$$

Now, for the given temperature, the value of most probable energy is given as:

$$\begin{gathered} k_p=\frac{1}{2}\left(8.62\times {10}^{-5}\frac{eV}{K}\right)\left(1.5\times {10}^{7}K\right) \\ =6.5\times {10}^{2}eV \\ =0.65keV \end{gathered}$$

Hence, the equation of the most probable speed is verified and the value of probable energy at the given temperature is$0.65keV$.

b) Calculate the most probable speed

Equation (ii) gives the most probable speed in terms of the molar massM, and indicates its derivation. Since the massmof the particle is related toMby the Avogadro constant, then, the most probable speed can be given using equation (ii) and equation (iii) as:

$$\begin{gathered} v_P=\sqrt{\frac{2RT}{M}} \\ =\sqrt{\frac{2RT}{m{N}_A}}\left(\because fromequation\left(ii\right)\right) \\ =\sqrt{\frac{2kT}{m}} \end{gathered}$$ …… (a)

Now, for the given mass of proton, the most probable speed of a proton can be given as:

$$\begin{gathered} v_p=\sqrt{\frac{2\left(1.38\times {10}^{-23}{\displaystyle \frac{J}{K}}\right)\left(1.5\times {10}^{7}K\right)}{\left(1.67\times {10}^{-27}kg\right)}} \\ =5.0\times {10}^5\frac{m}{s} \end{gathered}$$

Hence, the equation of the most probable speed is verified and the value of the probable speed of a proton is $5.0\times {10}^5\frac{m}{s}$.

c) Calculate the corresponding kinetic energy

The corresponding kinetic energy to the most probable speed can be given by substituting equation (a) in equation (iv) as follows:

$$\begin{gathered} k_{v,p}=\frac{1}{2}m{\left(\sqrt{\frac{2kT}{m}}\right)}^2 \\ =\frac{m}{2}\left(\frac{2kT}{m}\right) \\ =kT \end{gathered}$$

Now, the value of the kinetic energy at the given temperature is given as follows:

$$\begin{gathered} k_{v,p}=\left(8.62\times {10}^{-5}\frac{eV}{K}\right)\left(1.5\times {10}^{7}K\right) \\ =1.3\times {10}^{3}eV \\ =1.3keV \end{gathered}$$

Hence, the equation of the corresponding energy is verified that is which is indicated in Fig. 43-10 by a single vertical line and the value of the kinetic energy at the given temperature isrole="math" localid="1661933445435" $1.3\mathrm{ke}V$
.