When most elastomeric polymers (e.g., a rubber band) are stretched, the molecules become more ordered, as illustrated here: Suppose you stretch a rubber band. (a) Do you expect the entropy of the system to increase or decrease? (b) If the rubber band were stretched isothermally, would heat need to be absorbed or emitted to maintain constant temperature? (c) Try this experiment: Stretch a rubber band and wait a moment. Then place the stretched rubber band on your upper lip, and let it return suddenly to its unstretched state (remember to keep holding on!). What do you observe? Are your observations consistent with your answer to part (b)?

Short Answer

Expert verified
When a rubber band is stretched, its molecules become more ordered, leading to a decrease in entropy. If stretched isothermally, heat must be absorbed to maintain constant temperature. When a stretched rubber band returns to its unstressed state, it releases the absorbed heat, feeling cool to the touch. This observation is consistent with the requirement of heat absorption during isothermal stretching.

Step by step solution

01

(a) Entropy Change

When a rubber band is stretched, its molecules become more ordered as opposed to their relatively random alignment in the unstressed state. Entropy is a measure of the randomness or disorder of a system. So, when the rubber band is stretched and its molecules become more ordered, the entropy of the system decreases.
02

(b) Heat Absorption/Emission

If the rubber band is stretched isothermally, it means the process takes place at a constant temperature. By stretching the rubber band, we are doing work on it, which requires energy. Since the entropy of the system decreases in this process (from part (a)), according to the second law of thermodynamics, the heat must flow into the system to maintain the total entropy constant. Therefore, heat needs to be absorbed by the rubber band to maintain a constant temperature during an isothermal stretch.
03

(c) Observation after the Experiment

Stretch a rubber band and wait for a moment. Then, place the stretched rubber band on your upper lip and let it return suddenly to its unstressed state while still holding on to it. You will observe that the rubber band feels cool on your lip. This is because the rubber band absorbed heat from its surroundings while stretching (part (b)). When the rubber band returns to its unstressed state, the absorbed heat is released, making the rubber band feel cool to the touch. This observation is consistent with the answer given in part (b).

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Most popular questions from this chapter

(a) For a process that occurs at constant temperature, does the change in Gibbs free energy depend on changes in the enthalpy and entropy of the system? (b) For a certain process that occurs at constant \(T\) and \(P\), the value of \(\Delta G\) is positive. Is the process spontaneous? (c) If \(\Delta G\) for a process is large, is the rate at which it occurs fast?

In each of the following pairs, which compound would you expect to have the higher standard molar entropy: (a) \(\mathrm{C}_{3} \mathrm{H}_{\mathrm{s}}(g)\) or $\mathrm{C}_{4} \mathrm{H}_{10}(g)$, (b) \(\mathrm{C}_{4} \mathrm{H}_{10}(l)\) or \(\mathrm{C}_{4} \mathrm{H}_{10}(g)\)

Using the data in Appendix \(C\) and given the pressures listed, calculate \(K_{\mathrm{p}}\) and \(\Delta G\) for each of the following reactions: $$ \begin{array}{l} \text { (a) } \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) \\ \quad R_{\mathrm{N}_{2}}=263.4 \mathrm{kPa}, P_{\mathrm{H}_{2}}=597.8 \mathrm{kPa}, P_{\mathrm{NH}_{3}}=101.3 \mathrm{kPa} \\ \text { (b) } 2 \mathrm{~N}_{2} \mathrm{H}_{4}(g)+2 \mathrm{NO}_{2}(g) \longrightarrow 3 \mathrm{~N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g) \end{array} $$ \(P_{\mathrm{N}_{2} \mathrm{H}_{4}}=P_{\mathrm{NO}_{2}}=5.07 \mathrm{kPa}\) $$ \begin{array}{l} \quad R_{\mathrm{N}_{2}}=50.7 \mathrm{kPa}, P_{\mathrm{H}_{2} \mathrm{O}}=30.4 \mathrm{kPa} \\ \text { (c) } \mathrm{N}_{2} \mathrm{H}_{4}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g) \\ P_{\mathrm{N}_{2} \mathrm{H}_{4}}=101.3 \mathrm{kPa}, P_{\mathrm{N}_{2}}=152.0 \mathrm{kPa}, P_{\mathrm{H}_{2}}=253.3 \mathrm{kPa} \end{array} $$

The standard entropies at \(298 \mathrm{~K}\) for certain group 14 elements are: \(\mathrm{C}(s,\) diamond $)=2.43 \mathrm{~J} / \mathrm{mol}-\mathrm{K}, \mathrm{Si}(s)=18.81 \mathrm{~J} /$ $\mathrm{mol}-\mathrm{K}, \mathrm{Ge}(s)=31.09 \mathrm{~J} / \mathrm{mol}-\mathrm{K}, \quad\( a n d \)\quad \mathrm{Sn}(s)=51.818 \mathrm{~J} /$ mol-K. All but \(S\) n have the same (diamond) structure. How do you account for the trend in the \(S^{\circ}\) values?

Carbon disulfide \(\left(C S_{2}\right)\) is a toxic, highly flammable substance. The following thermodynamic data are available for \(\mathrm{CS}_{2}(I)\) and \(\mathrm{CS}_{2}(g)\) at \(298 \mathrm{~K}\) \begin{tabular}{lcc} \hline & \(\Delta H_{i}(\mathrm{k} / \mathrm{mol})\) & $\Delta G_{i}^{\prime}(\mathrm{kJ} / \mathrm{mol})$ \\ \hline\(C S_{2}(l)\) & 89.7 & 65.3 \\ \(C S_{2}(g)\) & 117.4 & 67.2 \\ \hline \end{tabular} (a) Draw the Lewis structure of the molecule. What do you predict for the bond order of the \(\mathrm{C}-\mathrm{S}\) bonds? \((\mathbf{b})\) Use the VSEPR method to predict the structure of the \(\mathrm{CS}_{2}\) molecule. (c) Liquid \(\mathrm{CS}_{2}\) burns in \(\mathrm{O}_{2}\) with a blue flame, forming \(\mathrm{CO}_{2}(g)\) and \(\mathrm{SO}_{2}(g)\). Write a balanced equation for this reaction. (d) Using the data in the preceding table and in Appendix \(C,\) calculate \(\Delta H^{\circ}\) and \(\Delta G^{\circ}\) for the reaction in part \((c) .\) Is the reaction exothermic? Is it spontaneous at \(298 \mathrm{~K} ?\) (e) Use the data in the table to calculate \(\Delta S^{\circ}\) at $298 \mathrm{~K}\( for the vaporization of \)\mathrm{CS}_{2}(I) .$ Is the sign of \(\Delta S^{\circ}\) as you would expect for a vaporization? (f) Using data in the table and your answer to part (e), estimate the boiling point of \(\mathrm{CS}_{2}(l)\). Do you predict that the substance will be a liquid or a gas at \(298 \mathrm{~K}\) and \(101.3 \mathrm{kPa}\) ?

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