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Chapter 20: Problem 50

Suppose \(0.550 \mathrm{~mol}\) of an ideal gas is isothermally and reversibly expanded in the four situations given below. What is the change in the entropy of the gas for each situation? $$ \begin{array}{lcccl} \hline \text { Situation } & \text { (a) } & \text { (b) } & \text { (c) } & \text { (d) } \\ \hline \text { Temperature (K) } & 250 & 350 & 400 & 450 \\ \text { Initial volume }\left(\mathrm{cm}^{3}\right) & 0.200 & 0.200 & 0.300 & 0.300 \\ \text { Final volume }\left(\mathrm{cm}^{3}\right) & 0.800 & 0.800 & 1.20 & 1.20 \\ \hline \end{array} $$

Short Answer

Expert verified
In each situation, the entropy change is approximately 8.779 J/K.

Step by step solution

01

Understanding the Formula for Entropy Change

For an isothermal, reversible expansion of an ideal gas, the change in entropy \( \Delta S \) can be calculated using the formula:\[\Delta S = nR \ln \left( \frac{V_f}{V_i} \right)\]where \( n \) is the number of moles, \( R \) is the universal gas constant \( 8.314 \text{ J/mol路K} \), \( V_f \) is the final volume, and \( V_i \) is the initial volume. In this problem \( n = 0.550 \text{ mol} \).
02

Situation (a) Calculations

For situation (a), the temperature = 250 K, initial volume \( V_i = 0.200 \text{ L} \), final volume \( V_f = 0.800 \text{ L} \).Calculate the change in entropy:\[\Delta S_a = 0.550 \times 8.314 \ln \left( \frac{0.800}{0.200} \right)\]\[\Delta S_a = 0.550 \times 8.314 \ln (4)\]\[\Delta S_a \approx 6.332 \times 1.386 = 8.779 \text{ J/K}\]
03

Situation (b) Calculations

For situation (b), the temperature = 350 K, initial volume \( V_i = 0.200 \text{ L} \), final volume \( V_f = 0.800 \text{ L} \).\[\Delta S_b = 0.550 \times 8.314 \ln \left( \frac{0.800}{0.200} \right)\]This calculation is similar to situation (a) because the volume change is the same, therefore:\[\Delta S_b \approx 8.779 \text{ J/K}\]
04

Situation (c) Calculations

For situation (c), the temperature = 400 K, initial volume \( V_i = 0.300 \text{ L} \), final volume \( V_f = 1.20 \text{ L} \).\[\Delta S_c = 0.550 \times 8.314 \ln \left( \frac{1.20}{0.300} \right)\]\[\Delta S_c = 0.550 \times 8.314 \ln (4)\]\[\Delta S_c \approx 8.779 \text{ J/K}\]
05

Situation (d) Calculations

For situation (d), the temperature = 450 K, initial volume \( V_i = 0.300 \text{ L} \), final volume \( V_f = 1.20 \text{ L} \).Following the same process:\[\Delta S_d = 0.550 \times 8.314 \ln \left( \frac{1.20}{0.300} \right)\]\[\Delta S_d \approx 8.779 \text{ J/K}\]
06

Summarizing Results

The change in entropy for each situation (a, b, c, d) is approximately the same since they have identical volume ratios:- \( \Delta S_a \approx 8.779 \text{ J/K} \)- \( \Delta S_b \approx 8.779 \text{ J/K} \)- \( \Delta S_c \approx 8.779 \text{ J/K} \)- \( \Delta S_d \approx 8.779 \text{ J/K} \)The calculated entropy changes do not depend on temperature in these scenarios as they involve the same volume expansion ratio for each case.

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Key Concepts

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

Entropy Change
Entropy is a measure of disorder or randomness in a system. In the context of thermodynamics, entropy change (abla S) specifically refers to the change in entropy as a system transitions from one state to another.
For an isothermal (constant temperature) and reversible process involving an ideal gas, we use the formula:
  • \[\Delta S = nR \ln \left( \frac{V_f}{V_i} \right)\]
  • where \( n \) is the number of moles, \( R \) is the universal gas constant (8.314 J/mol路K), \( V_f \) is the final volume, and \( V_i \) is the initial volume.
In all the situations provided, the entropy change is dictated by the relative change in volume. If the volume ratio, \( \frac{V_f}{V_i} \), remains constant, the entropy change will also remain unaffected by the temperature.
Ideal Gas
An ideal gas is a simplified model of a gas where the molecules do not interact with each other except by perfectly elastic collisions. It's an important concept in thermodynamics as it helps in deriving formulas that model how real gases behave under many conditions.
The behavior of an ideal gas is described by the Ideal Gas Law:
  • \[pV = nRT\]
  • where \( p \) is pressure, \( V \) is volume, \( n \) is number of moles, \( R \) is the universal gas constant, and \( T \) is temperature.
This equation relates the primary physical properties of an ideal gas. In our exercise, knowing that the gas is ideal allows us to focus on its volume change without worrying about molecular interactions. This simplifies our calculations, allowing us to use clean and straightforward mathematical expressions to predict changes such as entropy.
Isothermal Process
An isothermal process is a thermodynamic process that occurs at a constant temperature. In such scenarios, any heat added to the system is used to perform work rather than changing the temperature.
For isothermal transformations involving an ideal gas, the relationship between pressure and volume remains a simple inverse relationship dominated by Boyle's Law:
  • \[pV = ext{constant}\]
Because the temperature doesn't change, the internal energy of the ideal gas does not change either. This is significant because it simplifies the analysis of energy interactions, focusing solely on mechanical work and the accompanying entropy changes.
Reversible Expansion
Reversible expansion in thermodynamics implies a system that can be reversed at any step without leaving traces. It is a theoretical construct allowing maximum efficiency flow of energy in systems.
Throughout an ideal gas's reversible expansion, each incremental change is just balanced by external forces, making it quasi-static. In this context, the formula for entropy change closely follows the logarithmic volume relationship outlined earlier:
  • A system that can expand and shrink without friction or loss optimizes energy transfer.
  • The calculated entropy change in our problems illustrates the maximum possible entropy shift during such ideal expansions.
Thus, reversibility ensures that the process follows the path of equilibrium, reinforcing why entropy calculations in our exercise consistently reach the same values across differing conditions.

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Most popular questions from this chapter

Suppose \(1.0 \mathrm{~mol}\) of a monatomic ideal gas initially at \(10 \mathrm{~L}\) and \(300 \mathrm{~K}\) is heated at constant volume to \(600 \mathrm{~K}\), allowed to expand isothermally to its initial pressure, and finally compressed at constant pressure to its original volume, pressure, and temperature. During the cycle, what are (a) the net energy entering the system (the gas) as heat and (b) the net work done by the gas? (c) What is the efficiency of the cycle?

The electric motor of a heat pump transfers energy as heat from the outdoors, which is at \(-5.0^{\circ} \mathrm{C}\), to a room that is at \(17^{\circ} \mathrm{C}\). If the heat pump were a Carnot heat pump (a Carnot engine working in reverse), how much energy would be transferred as heat to the room for each joule of electric energy consumed?

The efficiency of a particular car engine is \(25 \%\) when the engine does \(8.2 \mathrm{~kJ}\) of work per cycle. Assume the process is reversible. What are (a) the energy the engine gains per cycle as heat \(Q_{\text {gain }}\) from the fuel combustion and (b) the energy the engine loses per cycle as heat \(Q_{\text {lost }} ?\) If a tune-up increases the efficiency to \(31 \%\), what are (c) \(Q_{\text {gain }}\) and (d) \(Q_{\text {lost }}\) at the same work value?

(a) During each cycle, a Carnot engine absorbs \(750 \mathrm{~J}\) as heat from a high-temperature reservoir at \(360 \mathrm{~K}\), with the low-temperature reservoir at \(280 \mathrm{~K}\). How much work is done per cycle? (b) The engine is then made to work in reverse to function as a Carnot refrigerator between those same two reservoirs. During each cycle, how much work is required to remove \(1200 \mathrm{~J}\) as heat from the low-temperature reservoir?

An insulated Thermos contains \(130 \mathrm{~g}\) of water at \(80.0^{\circ} \mathrm{C}\). You put in a \(12.0 \mathrm{~g}\) ice cube at \(0^{\circ} \mathrm{C}\) to form a system of ice \(+\) original water. (a) What is the equilibrium temperature of the system? What are the entropy changes of the water that was originally the ice cube (b) as it melts and (c) as it warms to the equilibrium temperature? (d) What is the entropy change of the original water as it cools to the equilibrium temperature? (e) What is the net entropy change of the ice \(+\) original water system as it reaches the equilibrium temperature?
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