Question

Figure 12.57 shows a one-dimensional row of 5 microscopic objects each of mass $\mathbf{4}\mathbf{.}{\mathbf{10}}^{\mathbf{-}\mathbf{26}}\mathbf{kg}$, connected by forces that can be modeled by springs of stiffness 15 N/m. These objects can move only along the x axis.

(a) Using the Einstein model, calculate the approximate entropy of this system for total energy of 0, 1, 2, 3, 4, and 5 quanta. Think carefully about what the Einstein model is, and apply those concepts to this one-dimensional situation. (b) Calculate the approximate temperature of the system when the total energy is 4 quanta. (c) Calculate the approximate specific heat on a per-object basis when the total energy is 4 quanta. (d) If the temperature is raised very high, what is the approximate specific heat on a per-object basis? Give a numerical value and compare with your result in part (c).

Short answer

1. The entropy ofsystem for total energy of 0, 1, 2, 3, 4, and 5 quanta is listed in the table below.
2. The temperature of the energy 4 quanta is 462 K.
3. The specific heat of 4 quanta of energy is$1.10\times {10}^{-23}\mathrm{J}/\mathrm{K}$.
4. The specific heat on per object is which is obviously higher than obtained in part c but the order is same.

Step-by-step solution

Identification of given data

The given data can be listed below,

• The number of microstates is, N=5.
• The mass of each object is, $m=4\times {10}^{-26}\mathrm{kg}$.
• The stiffness of the spring is, $k=15\mathrm{N}/\mathrm{m}$.

Concept/Significance of entropy of the system

The entropy of a thing is a measurement of how much energy is unavailable for work. The entropy also refers to the number of potential atom combinations in a structure. Entropy is a measure of randomness and variation.

Determination of the approximate entropy of the system for total energy of 0, 1, 2, 3, 4, and 5 quanta

The effective stiffness of spring is given by,

$k_{eff}=4k$

The entropy of an object is given by,

localid="1657862994876" $S\mathit{=}{k}_B\mathrm{In\Omega }$

Here, $\Omega$is the number of microstates whose value islocalid="1657862997799" $\Omega =\frac{\left(q+N-1\right)!}{q!\left(N-1\right)!}$

The entropy of each the quanta of energy is given in table below,

(b) Determination of the approximate temperature of the system when the total energy is 4 quanta.

The energy of the system is given by,

$E=\sout{h}\omega$

Here, is the Planck constant, and is the frequency whose value is $\sqrt{\frac{k_{eff}}{m}}$

Substitute all the values in the above,

$\mathrm{E}=\left(1.05\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)\sqrt{\frac{4\times 15\mathrm{N}/\mathrm{m}}{4\times {10}^{-26}\mathrm{kg}}}$

$=4.08\times {10}^{-21}\mathrm{J}$

The temperature of system is given by,

$T=\frac{∆E}{∆S}$

Temperature of the 4 quanta is given by,

$T=\frac{5\sout{h}\omega -3\sout{h}\omega }{S\left(5\right)-S\left(3\right)}$

Substitute all the values in the above equation.

$$\begin{gathered} T=\frac{\left(5-3\right)\left(4.08\times {10}^{-21}\mathrm{J}\right)}{6.67\times {10}^{-23}\mathrm{J}/\mathrm{K}-4.91\times {10}^{-23}\mathrm{J}/\mathrm{K}} \\ =462\mathrm{K} \end{gathered}$$

Thus, the temperature of the energy 4 quanta is $462\mathrm{K}$.

(c) Determination of the approximate specific heat on a per-object basis when the total energy is 4 quanta.

The specific heat of 4 quanta is given by,

$$\begin{gathered} c\left(4\right)=\frac{1}{N}\frac{∆E}{∆T} \\ =\frac{1}{N}\frac{\sout{h}\omega }{T\left(4.5\right)-T\left(3.5\right)} \end{gathered}$$

The temperature of 3.5 quanta is given by,

$T\left(3.5\right)=\frac{\sout{h}\omega }{S\left(4\right)-T\left(3\right)}$

Substitute all the values in the above,

$$\begin{gathered} \mathrm{T}\left(3.5\right)=\frac{4.08\times {10}^{-21}}{5.86\times {10}^{-23}\mathrm{J}/\mathrm{K}-4.91\times {10}^{-23}\mathrm{J}/\mathrm{K}} \\ =429\mathrm{K} \end{gathered}$$

The temperature of 4.5 quanta is given by,

$T\left(4.5\right)=\frac{\sout{h}\omega }{S\left(5\right)-S\left(4\right)}$

Substitute all the values in the above,

$$\begin{gathered} \mathrm{T}\left(4.5\right)=\frac{4.08\times {10}^{-21}\mathrm{J}}{6.67\times {10}^{-23}\mathrm{J}/\mathrm{K}-5.86\times {10}^{-23}\mathrm{J}/\mathrm{K}} \\ =503\mathrm{K} \end{gathered}$$

Substitute all the values in the specific heat equation.

$$\begin{gathered} \mathrm{c}\left(4\right)=\frac{4.08\times {10}^{-21}\mathrm{J}}{5\left(503\mathrm{K}-429\mathrm{K}\right)} \\ =1.10\times {10}^{-23}\mathrm{J}/\mathrm{K} \end{gathered}$$

Thus, the specific heat of 4 quanta of energy is $1.10\times {10}^{-23}\mathrm{J}/\mathrm{K}$.

(d) Determination of the approximate specific heat on a per-object and comparison with answer obtain in c part.

The heat capacity is three times of the Boltzmann constant for higher order of energy that is given by,

$c=3k_B$

Substitute values in the above,

$$\begin{gathered} \mathrm{c}=3\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right) \\ =4.14\times {10}^{-23}\mathrm{J}/\mathrm{K} \end{gathered}$$

Thus, the specific heat on per object is$=4.14\times {10}^{-23}\mathrm{J}/\mathrm{K}$ which is obviously higher than obtained in part c but the order is same.