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Chapter 19: Problem 2

As shown here, one type of computer keyboard cleaner contains liquefied 1,1 -difluoroethane \(\left(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{~F}_{2}\right),\) which is a gas at atmospheric pressure. When the nozzle is squeezed, the 1,1 -difluoroethane vaporizes out of the nozzle at high pressure, blowing dust out of objects. (a) Based on your experience, is the vaporization a spontaneous process at room temperature? (b) Defining the 1,1 -difluoroethane as the system, do you expect \(q_{\text {sys }}\) for the process to be positive or negative? (c) Predict whether \(\Delta S\) is positive or negative for this process. (d) Given your answers to \((a),(b),\) and \((c),\) do you think the operation of this product depends more on enthalpy or entropy? [Sections 19.1 and 19.2\(]\)

Short Answer

Expert verified
(a) The vaporization of 1,1-difluoroethane from the computer keyboard cleaner is a spontaneous process at room temperature, as it occurs naturally and spontaneously when the nozzle is squeezed. (b) As the 1,1-difluoroethane vaporizes, it absorbs heat from the surroundings, resulting in a positive \(q_{\text{sys}}\). (c) The change in entropy (\(\Delta S\)) for the process is positive, as the system goes from a more ordered state (liquid) to a less ordered state (gas). (d) The operation of the keyboard cleaner depends more on entropy than enthalpy, as the increasing disorder in the system due to the transformation of liquid into gas is the major driving factor in the process.

Step by step solution

01

(Discuss the vaporization process spontaneity)

At room temperature, when we squeeze the nozzle, the 1,1-difluoroethane vaporizes out of the nozzle at high pressure. Based on daily experience with aerosol sprays, we know that this vaporization comes naturally and spontaneously once we apply pressure, meaning the process is spontaneous at room temperature.
02

(Predict the value of \(q_{sys}\))

Since the 1,1-difluoroethane vaporizes and turns into gas, it is absorbing heat from the surroundings. Therefore, we should expect the heat absorbed by the system (\(q_{sys}\)) to be positive in this process.
03

(Predict the sign of \(ΔS\))

When the liquefied 1,1-difluoroethane vaporizes, it goes from a more ordered state (liquid) to a less ordered state (gas). The disorder increases as molecules become more free to move in the gaseous state than in the liquid state. As a result, we can conclude that the change in entropy (\(\Delta S\)) for this process is positive.
04

(Analyze if the operation depends on enthalpy or entropy)

Given the spontaneous process, positive \(q_{sys}\), and positive \(\Delta S\), we can deduce that the operation has both favorable enthalpy and entropy changes. However, since the vaporization process mainly includes the transformation of a liquid into a gas, the entropy change is the major driving factor in this process, attributing to the increasing disorder in the system. Thus, it's likely that the operation of this product depends more on entropy than enthalpy.

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Most popular questions from this chapter

Indicate whether each statement is true or false. (a) \(\Delta S\) depends on whether the process is reversible or irreversible. \((\mathbf{b})\) If a system undergoes an irreversible change, the entropy of the universe increases. (c) Only if the change in entropy of the system is exactly matched by an equal and opposite change in the entropy of the surroundings, the system undergoes a reversible process. (d) If the entropy change of the system is zero, the system undergoes a reversible process.

The oxidation of glucose $\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\( in body tissue produces \)\mathrm{CO}_{2}$ and \(\mathrm{H}_{2} \mathrm{O} .\) In contrast, anaerobic decomposition, which occurs during fermentation, produces ethanol $\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\( and \)\mathrm{CO}_{2}$. (a) Using data given in Appendix \(\mathrm{C}\), compare the equilibrium constants for the following reactions: $$ \begin{array}{r} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \rightleftharpoons 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s) \rightleftharpoons 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(I)+2 \mathrm{CO}_{2}(g) \end{array} $$ (b) Compare the maximum work that can be obtained from these processes under standard conditions.

The crystalline hydrate $\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2} \cdot 4 \mathrm{H}_{2} \mathrm{O}(s)$ loses water when placed in a large, closed, dry vessel at room temperature: $$ \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2} \cdot 4 \mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}(s)+4 \mathrm{H}_{2} \mathrm{O}(g) $$ This process is spontaneous and \(\Delta H^{\circ}\) is positive at room temperature. (a) What is the sign of \(\Delta S^{\circ}\) at room temperature? (b) If the hydrated compound is placed in a large, closed vessel that already contains a large amount of water vapor, does \(\Delta S^{\circ}\) change for this reaction at room temperature?

For each of the following pairs, predict which substance has the higher entropy per mole at a given temperature: (a) \(\mathrm{I}_{2}(s)\) or \(\mathrm{I}_{2}(g)\) (b) \(\mathrm{O}_{2}(g)\) at \(50.7 \mathrm{kPa}\) or \(\mathrm{O}_{2}\) at \(101.3 \mathrm{kPa}\) (c) 1 molof \(\mathrm{N}_{2}\) in 22.4 Lor \(1 \mathrm{~mol}\) of \(\mathrm{N}_{2}\) in \(44.8 \mathrm{~L}\). (d) \(\mathrm{CH}_{3} \mathrm{OH}(I)\) or \(\mathrm{CH}_{3} \mathrm{OH}(s)\)

(a) What sign for \(\Delta S\) do you expect when the pressure on 0.600 mol of an ideal gas at \(350 \mathrm{~K}\) is increased isothermally from an initial pressure of \(76.0 \mathrm{kPa} ?(\mathbf{b})\) If the final pressure on the gas is \(121.6 \mathrm{kPa}\), calculate the entropy change for the process. (c) Do you need to specify the temperature to calculate the entropy change?
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