Figure 20-33 gives the force magnitude Fversus stretch distance xfor a rubber band, with the scale of the Faxis set by${\mathit{F}}_{\mathbf{s}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{}\mathit{N}$and the scale of the x-axis set by${\mathit{x}}_{\mathbf{s}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{50}\mathit{c}\mathit{m}$ .The temperature is$\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{°}\mathit{C}$ . When the rubber band is stretched by$\mathit{x}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{70}\mathit{c}\mathit{m}$, at what rate does the entropy of the rubber band change during a small additional stretch?

The axis is set by force,$F_x=1.50\mathrm{N}$.

The axis is set by stretch,$x_s=3.50\mathrm{cm}\left(\frac{1\mathrm{cm}}{100\mathrm{cm}}\right)=0.035\mathrm{m}$.

The temperature,$T=2.00°\mathrm{C}=275\mathrm{K}$.

The additional stretch, $x=1.70\mathrm{cm}\left(\frac{1\mathrm{cm}}{100\mathrm{cm}}\right)=0.017\mathrm{m}$.

By using Hooke's law, we can find the value of the spring constant. By using Equations 20-6 and 20-7, we can find the change in entropy. By substituting the equation for work done and differentiating for x, we can find the rate of change of entropy of the rubber band during a small additional stretch.

Formula:

The force of a spring using Hooke’s law, ${\mathrm{F}}_{\mathrm{s}}={\mathrm{kx}}_{\mathrm{s}}$ (1)

where, $F_s$ is force of stretch, k is the spring constant, and $x_s$is the displacement.

The total entropy change of a gas from equation 20-7,

$$\begin{gathered} ∆S=∆{\mathrm{S}}_{\mathrm{H}}+∆{\mathrm{S}}_{\mathrm{L}} \\ =\frac{\left|{\mathrm{Q}}_{\mathrm{H}}\right|}{{\mathrm{T}}_{\mathrm{H}}}-\frac{\left|{\mathrm{Q}}_{\mathrm{L}}\right|}{{\mathrm{T}}_{\mathrm{L}}} \end{gathered}$$ (2)

The work done of a cycle,$W=\left|{\mathrm{Q}}_{\mathrm{H}}\right|-\left|{\mathrm{Q}}_{\mathrm{L}}\right|$ (3)

The work done on the spring due to stretch, $\mathrm{dW}=\mathrm{Fdx}$ (4)

The spring constant of the rubber band using equation (1) and given values is calculated as:

$$\begin{gathered} k=\frac{{\mathrm{S}}_{\mathrm{s}}}{{\mathrm{x}}_{\mathrm{s}}} \\ =\frac{1.50\mathrm{N}}{0.0350\mathrm{m}} \\ =42.86\mathrm{N}/\mathrm{m} \end{gathered}$$

We know that the work done is given by integrating the equation (4) as follows:

$$\begin{gathered} \int \mathrm{dW}=\int \mathrm{Fdx} \\ \mathrm{W}=\int \mathrm{Fdx} \\ \mathrm{W}=\frac{{\mathrm{kx}}^2}{2} \end{gathered}$$

Putting this in the equation (2) for,${\mathrm{T}}_{\mathrm{H}}\left(={\mathrm{T}}_{\mathrm{L}}\right)=\mathrm{T}$and using equation (3), we get the entropy change as given:

$$\begin{gathered} ∆S=\frac{\mathrm{W}}{\mathrm{T}} \\ =\frac{{\mathrm{kx}}^2}{2\mathrm{T}} \end{gathered}$$

Therefore, the rate of change of entropy with a small additional stretch is given by differentiating the above equation by dx using given values as given:

$$\begin{gathered} \frac{\mathrm{dS}}{\mathrm{dx}}=\frac{2\mathrm{kx}}{2\mathrm{T}} \\ \frac{\mathrm{dS}}{\mathrm{dx}}=\frac{\mathrm{kx}}{\mathrm{T}} \\ \left|\frac{\mathrm{dS}}{\mathrm{dx}}\right|=\frac{\mathrm{k}\left|\mathrm{x}\right|}{\mathrm{T}} \\ =\frac{42.86\mathrm{N}/\mathrm{m}\times 0.0170\mathrm{m}}{275\mathrm{K}} \\ =2.65\times {10}^{-3}\mathrm{J}/\mathrm{K}·\mathrm{m} \end{gathered}$$

Hence, the value of the rate of entropy change is $2.65\times {10}^{-3}\mathrm{J}/\mathrm{K}·\mathrm{m}$