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Chapter 19: Problem 28

The pressure on \(0.850\) mol of neon gas is increased from \(1.25\) atm to \(2.75\) atm at \(100^{\circ} \mathrm{C}\). Assuming the gas to be ideal, calculate \(\Delta S\) for this process.

Short Answer

Expert verified
The change of entropy (\(\Delta S\)) for this isothermal process can be calculated using the formula \(\Delta S = nR * \ln(P2/P1)\), where n = 0.850 mol, R = 0.0821 L * atm / (mol * K), P1 = 1.25 atm, and P2 = 2.75 atm. Plugging in the values, we get \(\Delta S \approx 0.0550\) L * atm / K.

Step by step solution

01

Identify the given parameters

We are given the following parameters: - Number of moles (n) = 0.850 mol - Initial pressure (P1) = 1.25 atm - Final pressure (P2) = 2.75 atm - Temperature (T) = 100°C Since all calculations must be in Kelvin, we need to convert the temperature from Celsius to Kelvin: T = 100°C + 273.15 = 373.15 K
02

Use the formula for the entropy change of an ideal gas during an isothermal process

Now we'll use the formula for the entropy change during an isothermal process: ΔS = nR * ln(P2/P1) Where n = 0.850 mol, R = 0.0821 L * atm / (mol * K), P1 = 1.25 atm, and P2 = 2.75 atm.
03

Calculate the entropy change

Plug the values into the formula: ΔS = (0.850 mol) * (0.0821 L * atm / (mol * K)) * ln(2.75 atm / 1.25 atm) ΔS ≈ 0.0698 * ln(2.2) ΔS ≈ 0.0698 * 0.7885 ΔS ≈ 0.0550 L * atm / K The entropy change (ΔS) for this process is approximately 0.0550 L * atm / K.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ideal Gas
An ideal gas is a theoretical concept that represents a gas composed of many randomly moving particles that are not subject to any force except during elastic collisions. The ideal gas law, which is a cornerstone in understanding ideal gases, is expressed as:\[ PV = nRT \]Here, \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.
  • The concept of an ideal gas helps us model real gases under many conditions though no real gas perfectly fits these criteria.
  • Neon, like other noble gases, behaves very closely to an ideal gas at standard temperature and pressure.
Even though ideal gases are a simplification, using this model allows us to predict the behavior of gases under various situations. It is particularly useful in calculations involving entropy changes during different gas processes.
Isothermal Process
An isothermal process occurs at a constant temperature. In such a process, the system exchanges heat with its surroundings to maintain the temperature constant despite changes in volume and pressure. In the realm of gases, particularly ideal gases, the work done and the heat exchange during an isothermal process are intimately linked.
  • For an isothermal process involving an ideal gas, the change in internal energy is zero. This is due to the direct relation of internal energy with temperature, which remains unchanged.
  • Entropy change in an isothermal process can be calculated using the formula:\[ \Delta S = nR \ln\left( \frac{P_2}{P_1} \right) \]This equation reflects how entropy is influenced by the pressure ratio when the temperature holds steady.
Understanding isothermal processes is crucial for analyzing systems where temperature remains constant, such as in the given example that calculates the entropy change (\( \Delta S \)) of neon gas during such a process at 100°C.
Neon Gas
Neon gas, a member of the noble gas family, is famous for its lack of color, taste, and distinct chemical inertness due to its full electron shells. This makes neon excellent for uses such as in neon signs and high-voltage indicators.
  • Due to its monoatomic nature and noble gas characteristics, neon behaves almost ideally under standard conditions, which aids in applying the ideal gas law accurately.
  • In the condition described in the exercise, neon's properties allow for an effective demonstration of entropy calculations in isothermal processes for an ideal gas.
While neon is often used in visual technologies, its gaseous properties make it a great candidate for educational purposes in physics and chemistry, especially in demonstrating theoretical concepts like those involving entropy changes in ideal gases.

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Most popular questions from this chapter

For a certain chemical reaction, \(\Delta H^{\circ}=-35.4 \mathrm{~kJ}\) and \(\Delta S^{\circ}=-85.5 \mathrm{~J} / \mathrm{K} .(\mathrm{a}) \mathrm{ls}\) the reaction exothermic or en- dothermic? (b) Does the reaction lead to an increase or decrease in the randomness or disorder of the system? (c) Calculate \(\Delta G^{\circ}\) for the reaction at \(298 \mathrm{~K}\). (d) Is the reaction spontaneous at \(298 \mathrm{~K}\) under standard conditions?

Consider the reaction \(2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g) .\) (a) Using data from Appendix C, calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\). (b) Calculate \(\Delta G\) at \(298 \mathrm{~K}\) if the partial pressures of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) are \(0.40 \mathrm{~atm}\) and \(1.60 \mathrm{~atm}\), respectively.

The relationship between the temperature of a reaction, its standard enthalpy change, and the equilibrium constant at that temperature can be expressed as the following linear equation: $$ \ln K=\frac{-\Delta H^{\circ}}{R T}+\text { constant } $$ (a) Explain how this equation can be used to determine \(\Delta H^{\circ}\) experimentally from the equilibrium constants at several different temperatures. (b) Derive the preceding equation using relationships given in this chapter. To what is the constant equal?

Carbon disulfide \(\left(\mathrm{CS}_{2}\right)\) is a toxic, highly flam mable substance. The following thermodynamic data are available for \(\mathrm{CS}_{2}(l)\) and \(\mathrm{CS}_{2}(g)\) at \(298 \mathrm{~K}\) : \begin{tabular}{lrl} \hline & \(\Delta H_{f}^{\circ}(\mathbf{k J} / \mathrm{mol})\) & \(\Delta G_{f}^{0}(\mathbf{k J} / \mathrm{mol})\) \\ \hline \(\mathrm{CS}_{2}(l)\) & \(89.7\) & \(65.3\) \\ \(\mathrm{CS}_{2}(g)\) & \(117.4\) & \(67.2\) \\ \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? (b) Use the VSEPR method to predict the structure of the \(\mathrm{CS}_{2}\) molecule. (c) Liquid \(\mathrm{CS}_{2}\) bums 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 K? (e) Use the data in the preceding table to calculate \(\Delta S^{\circ}\) at \(298 \mathrm{~K}\) for the vaporization of \(\mathrm{CS}_{2}(l)\). Is the sign of \(\Delta S^{\circ}\) as you would expect for a vaporization? (f) Using data in the preceding table and your answer to part (e), estimate the boiling point of \(\mathrm{CS}_{2}(\mathrm{l})\). Do you predict that the substance will be a liquid or a gas at \(298 \mathrm{~K}\) and \(1 \mathrm{~atm}\) ?

Aceticacid can be manufactured by combining methanol with carbon monoxide, an example of a carbonylation reaction: $$ \mathrm{CH}_{3} \mathrm{OH}(l)+\mathrm{CO}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{COOH}(l) $$ (a) Calculate the equilibrium constant for the reaction at \(25^{\circ} \mathrm{C}\). (b) Industrially, this reaction is run at temperatures above \(25^{\circ} \mathrm{C}\). Will an increase in temperature produce an increase or decrease in the mole fraction of acetic acid at equilibrium? Why are elevated temperatures used? (c) At what temperature will this reaction have an equilibrium constant equal to \(1 ?\) (You may assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are temperature independent, and you may ignore any phase changes that might occur.)
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