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20.36 . A lonely party balloon with a volume of 2.40 \(\mathrm{L}\) and containing 0.100 \(\mathrm{mol}\) of air is left behind to drift in the temporarily uninhabited and depressurized Intermational Space Station. Sunlight coming through a porthole heats and explodes the balloon, causing the air in it to undergo a free expansion into the empty station, whose total volume is 425 \(\mathrm{m}^{3}\) . Calculate the entropy change of the air during the expansion.

Short Answer

Expert verified
The entropy change is approximately 10.04 J/K.

Step by step solution

01

Identify key variables

We are given the initial volume of the balloon as \( V_i = 2.40 \text{ L} = 0.00240 \text{ m}^3 \), since 1 \, \text{L} = 0.001 \, \text{m}^3. The final volume after expansion is given as \( V_f = 425 \, \text{ m}^3 \). The number of moles of gas in the balloon is \( n = 0.100 \, \text{mol} \).
02

Use the entropy change for free expansion

In a free expansion, the external pressure is 0, so the energy transferred as work is also 0. For an ideal gas, the entropy change for a free expansion can be calculated using the equation:\[\Delta S = nR \ln\left(\frac{V_f}{V_i}\right)\]where \( R = 8.314 \, \text{J/(mol K)} \) is the ideal gas constant.
03

Calculate the volume ratio

Calculate the ratio \( \frac{V_f}{V_i} \) as follows:\[\frac{V_f}{V_i} = \frac{425}{0.00240} = 177083.33\]
04

Compute the entropy change

Substitute the values into the entropy change formula:\[\Delta S = 0.100 \, \text{mol} \times 8.314 \, \text{J/(mol K)} \times \ln(177083.33)\]Calculate the natural logarithm and then the entropy change:\[\ln(177083.33) \approx 12.084\]\[\Delta S = 0.100 \, \text{mol} \times 8.314 \, \text{J/(mol K)} \times 12.084 \approx 10.04 \, \text{J/K}\]
05

Conclusion

The entropy change of the air in the balloon during the free expansion is calculated to be approximately \( 10.04 \, \text{J/K} \).

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Key Concepts

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

Ideal Gas Law
The ideal gas law is a fundamental concept in chemistry and physics. It relates the pressure, volume, temperature, and amount of a gas. The formula is expressed as \( PV = nRT \), where:
  • \( P \) stands for the pressure of the gas.
  • \( V \) is the volume of the gas.
  • \( n \) represents the number of moles of the gas.
  • \( R \) is the ideal gas constant, equal to 8.314 J/(mol K).
  • \( T \) is the temperature in Kelvins.
This equation helps us predict how gases will react under different conditions. When applied in calculations, it assumes the gas behaves ideally, which means particles do not interact with each other and occupy no volume.
Although real gases may deviate slightly under high pressures or low temperatures, the ideal gas law provides a close approximation.It's a useful tool for understanding gas behavior, especially in scenarios like this problem, involving an air balloon inside the International Space Station.
Free Expansion
Free expansion is a process where a gas expands into a vacuum or an area of no pressure, without doing work on the environment. In this scenario, the gas doesn't exert force against any opposing forces or pressures.
Due to this lack of external pressure, free expansion doesn't involve energy transfer as work. The initial energy of the gas is solely retained within its particles, and only the volume changes.
This makes free expansion a key concept in thermodynamics, especially when analyzing entropy changes. The increased volume causes increased disorder among particles, which corresponds to higher entropy. The entropy change in this type of process for an ideal gas can be calculated using the formula:\[\Delta S = nR \ln\left(\frac{V_f}{V_i}\right)\]Here, the entropy change \( \Delta S \) arises due to the multiplication factor involving the ratio between final and initial volumes, \( V_f \) and \( V_i \). Understanding free expansion is essential for predicting entropy changes in thermodynamic transformations without external influences.
Thermodynamics
Thermodynamics is the branch of physics that deals with the relationships between heat, work, temperature, and energy. It defines how these elements interact in a system.
Key laws govern thermodynamics, including:
  • The First Law, which states that energy cannot be created or destroyed, only transformed.
  • The Second Law, which introduces the concept of entropy, indicating that in any energy transfer, some energy becomes unavailable to do work.
  • The Third Law, which suggests that as temperature approaches absolute zero, the entropy of a perfect crystal approaches a constant minimum.
In the case of the balloon in the exercise, thermodynamics helps us understand the change in entropy as the gas undergoes free expansion. This is guided by the principle that even in seemingly chaotic processes, nature trends towards equilibrium and increased entropy.
Thermodynamics provides a framework for understanding how energy transformations affect the state and properties of matter, making it crucial for solving problems related to changes in energy, such as the explosive expansion described.

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Most popular questions from this chapter

20.25. A sophomore with nothing better to do adds heat to 0.350 \(\mathrm{kg}\) of ice at \(0.0^{\circ} \mathrm{C}\) until it is all melted. (a) What is the change in entropy of the water? (b) The source of heat is a very massive body at a temperature of \(25.0^{\circ} \mathrm{C}\) . What is the change in entropy of this body? (c) What is the total change in entropy of the water and the heat source?

20.5. A certain nuclear-power plant has a mechanical-power output (used to drive an electric generator) of 330 \(\mathrm{MW}\) . Its rate of heat input from the nuclear reactor is 1300 \(\mathrm{MW}\) . (a) What is the thermal efficiency of the system? (b) At what rate is heat discarded by the system?

An aircraft engine takes in 9000 \(\mathrm{J}\) of heat and discards 6400 \(\mathrm{J}\) each cycle. (a) What is the mechanical work output of the engine during one cycle? (b) What is the thermal efficiency of the engine?

20.50. A stirling-cycle Engine. the Otto cycle, except that the compression and expansion of the gas are done at constant temperature, not adiabatically as in the Otto cycle. The Stirling cycle is used in external combustion engines (in fact, burning fuel is not necessary; any way of producing a temperature difference will do -solar, geothermal, ocean temperature gradient, etc. \(.\) which means that the gas inside the cylinder is not used in the combustion process. Heat is supplied by burning fuel steadily outside the cylinder, instead of explosively inside the cylinder as in the Otto cycle. For this reason Stirling-cycle engines are quieter than Otto-cycle engines, since there are no intake and exhaust valves (a major source of engine noise). While small Stirling engines are used for a variety of purposes, Stiring engines for automobiles have not been successful because they are larger, heavier, and more expensive than conventional automobile engines. In the cycle, the working fluid goes through the following sequence of steps (Fig. 20.30\()\) : (i) Compressed isothermally at temperature \(T_{1}\) from the initial state \(a\) to state \(b\) , with a compression ratio \(r .\) (ii) Heated at constant volume to state \(c\) at temperature \(T_{2}\) . (iii) Expanded isothermally at \(T_{2}\) to state \(d\) . (iv) Cooled at constant volume back to the initial state \(a\) . Assume that the working fluid is \(n\) moles of an ideal gas (for which \(C_{V}\) is independent of temperature). (a) Calculate \(Q, W,\) and \(\Delta U\) for each of the processes \(a \rightarrow b, b \rightarrow c, c \rightarrow d,\) and \(d \rightarrow a\) . (b) In the Stirling cycle, the heat transfers in the processes \(b \rightarrow c\) and \(d \rightarrow a\) do not involve external heat sources but rather use regeneration: The same substance that transfers heat to the gas inside the cylinder in the process \(b \rightarrow c\) also absorbs heat back from the gas in the process \(d \rightarrow a\) . Hence the heat transfers \(Q_{b \rightarrow c}\) and \(Q_{d \rightarrow a}\) do not play a role in determining the efficiency of the engine. Explain this last statement by comparing the expressions for \(Q_{b \rightarrow c}\) and \(Q_{d \rightarrow a}\) calculated in part (a). (c) Calculate the efficiency of a Stirling-cycle engine in terms of the temperatures \(T_{1}\) and \(T_{2}\) . How does this compare to the efficiency of a Carnot-cycle engine operating between these same two temperatures? (Historically, the Stirling cycle was devised before the Carnot cycle.) Does this result violate the second law of thermodynamics? Explain. Unfortunately, actual Stirling-cycle engines cannot achieve this efficiency due to problems with the heat-transfer processes and pressure losses in the engine.

20.26. You decide to take a nice hot bath but discover that your thoughtless roommate has used up most of the hot water. You fill the tub with 270 \(\mathrm{kg}\) of \(30.0^{\circ} \mathrm{C}\) water and attempt to warm it further by pouring in 5.00 \(\mathrm{kg}\) of boiling water from the stove. (a) Is this a reversible or an imeversible process? Use physical reasoning to explain. (b) Calculate the final temperature of the bath water. (c) Calculate the net change in entropy of the system (bath water + boiling water), assuming no heat exchange with the air or the tub itself.

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