/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 A lonely party balloon with a vo... [FREE SOLUTION] | 91Ó°ÊÓ

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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 International 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 10.051 J/K.

Step by step solution

01

Understand the Concept

Entropy change associated with a free expansion of a gas can be calculated using the formula for the change in entropy for an isothermal process. In this case, the gas is expanding into a vacuum, so the entropy change can be directly calculated because the final volume is known.
02

Identify the Given Values

We have an initial volume \( V_i = 2.40 \, \text{L} \) and final volume \( V_f = 425 \, \text{m}^3 \). Note that we need to convert the initial volume from liters to cubic meters. \( 2.40 \, \text{L} = 2.40 \times 10^{-3} \, \text{m}^3 \). The number of moles \( n = 0.100 \, \text{mol} \).
03

Use the Entropy Change Formula

The entropy change \( \Delta S \) for the isothermal expansion of an ideal gas is given by the formula: \[\Delta S = nR \ln \left( \frac{V_f}{V_i} \right)\]where \( R = 8.314 \, \text{J/mol·K} \) is the ideal gas constant.
04

Plug in the Values

Substitute the given values into the formula:\[\Delta S = 0.100 \, \text{mol} \times 8.314 \, \text{J/mol·K} \times \ln \left( \frac{425 \, \text{m}^3}{2.40 \times 10^{-3} \, \text{m}^3} \right)\]
05

Calculate the Logarithm

Calculate the logarithmic term:\[\ln \left( \frac{425}{2.40 \times 10^{-3}} \right) = \ln (177083.333) \approx 12.089\]
06

Calculate the Entropy Change

Multiply through to find the entropy change:\[\Delta S = 0.100 \, \times \, 8.314 \, \times \, 12.089 \approx 10.051 \, \text{J/K}\]
07

Conclusion

The entropy change of the air during the free expansion is approximately 10.051 J/K.

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Key Concepts

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

Free Expansion
Free expansion is a fascinating concept in thermodynamics. It refers to the behavior of a gas that expands into an available vacuum without any external resistance. Imagine letting go of an inflated balloon in an empty room. The air inside quickly swells to fill the entire space without any restraint.

During a free expansion, there are a few key characteristics to remember:
  • No work is done by or on the gas.
  • The process happens without any exchange of heat with the surroundings.
  • The internal energy of the gas remains constant.
This makes free expansion an adiabatic process, as no heat enters or leaves the system. In our initial example, when the balloon bursts in space, the air undergoes free expansion to fill the vast vacuum of the space station.
Ideal Gas
The concept of an ideal gas simplifies the study of gases by assuming certain conditions that make calculations more straightforward. An ideal gas is a theoretical gas composed of a set of randomly-moving, point particles that interact only through elastic collisions.

Some essential assumptions of an ideal gas include:
  • The gas particles themselves have no volume. They don't take up space individually.
  • There are no attractive or repulsive forces between the particles.
  • All collisions between particles and with the walls of the container are perfectly elastic, meaning there's no loss of kinetic energy.
  • The average kinetic energy of the gas particles is directly proportional to the gas's temperature.
These assumptions sometimes limit the ideal gas law's application to real gases, especially under conditions of high pressure or low temperature. However, in our calculation of the entropy change, assuming the air behaves as an ideal gas simplifies the process.
Isothermal Process
An isothermal process is an essential thermodynamical transformation in which a system changes while maintaining a constant temperature. In terms of gases, during an isothermal process, the internal energy change is zero because the temperature does not change. But the gas can perform work or have work done on it, exchanging energy in the form of heat.

For the ideal gas undergoing an isothermal expansion, as in the case of the balloon, the temperature remains uniform even as the gas expands since any extra internal energy is balanced by heat transfer from the environment.

This unique characteristic makes it possible to use specific formulas, like the entropy formula, to analyze changes effectively during the process since the ideal gas law applies throughout the transformation.
Entropy Calculation
Entropy is a measure of randomness or disorder in a system. Entropy calculation, especially during an isothermal process, helps understand changes in thermal energy spreading.

The formula used to calculate the change in entropy (Delta S) for an isothermal expansion of an ideal gas is:
\[Delta S = nR \ln \left( \frac{V_f}{V_i} \right) \]

Where:
  • Delta S is the change in entropy.
  • n is the number of moles of the gas.
  • R is the ideal gas constant, 8.314 J/mol·K.
  • \(V_f\) is the final volume, and \(V_i\) is the initial volume.
Plugging the known values into this equation allowed us to calculate the entropy change from the expansion of the air from a small initial volume to the vast space of the station. Understanding this formula shows how expansion increases disorder by spreading the gas molecules across a larger volume, thus resulting in positive entropy change.

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Most popular questions from this chapter

A Gasoline Engine. A gasoline engine takes in \(1.61 \times\) \(10^{4} \mathrm{J}\) of heat and delivers 3700 \(\mathrm{J}\) of work per cycle. The heat is obtained by burning gasoline with a heat of combustion of \(4.60 \times 10^{4} \mathrm{J} / \mathrm{g} .(\mathrm{a})\) What is the thermal efficiency? (b) How much heat is discarded in each cycle? (c) What mass of fuel is burned in each cycle? (d) If the engine goes through 60.0 cycles per second, what is its power output in kilowatts? In horsepower?

Consider a Diesel cycle that starts (at point \(a\) in Fig. 20.7\()\) with air at temperature \(T_{a}\) The air may be treated as an ideal gas. (a) If the temperature at point \(c\) is \(T_{c}\) , derive an expression for the efficiency of the cycle in terms of the compression ratio \(r .\) (b) What is the efficiency if \(T_{a}=300 \mathrm{K}, T_{c}=950 \mathrm{K}, \gamma=1.40\) and \(r=21.0 ?\)

A Human Engine. You decide to use your body as a Carnot heat engine. The operating gas is in a tube with one end in your mouth (where the temperature is \(37.0^{\circ} \mathrm{C} )\) and the other end at the surface of your skin, at \(30.0^{\circ} \mathrm{C}\) . (a) What is the maximum efficiency of such a heat engine? Would it be a very useful engine? (b) Suppose you want to use this human engine to lift a 2.50 -kg box from the floor to a tabletop 1.20 \(\mathrm{m}\) above the floor. How much must you increase the gravitational potential energy, and how much heat input is needed to accomplish this? (c) If your favorite candy bar has 350 food calories \((1\) food calorie \(=4186 \mathrm{J})\) and 80\(\%\) of the food energy goes into heat, how many of these candy bars must you eat to lift the box in this way?

Heat Pump. A heat pump is a heat engine run in reverse. In winter it pumps heat from the cold air outside into the warmer air inside the building, maintaining the building at a comfortable temperature. In summer it pumps heat from the cooler air inside the building to the warmer air outside, acting as an air conditioner. (a) If the outside temperature in winter is \(-5.0^{\circ} \mathrm{C}\) and the inside temperature is \(17.0^{\circ} \mathrm{C}\) , how many joules of heat will the heat pump deliver to the inside for each joule of electrical energy used to run the unit, assuming an ideal Carnot cycle? (b) Suppose you have the option of using electrical resistance heating rather than a heat pump. How much electrical energy would you need in order to deliver the same amount of heat to the inside of the house as in part (a)? Consider a Carnot heat pump delivering heat to the inside of a house to maintain it at \(68^{\circ} \mathrm{F}\) . Show that the heat pump delivers less heat for each joule of electrical energy used to operate the unit as the outside temperature decreases. Notice that this behavior is opposite to the dependence of the efficiency of a Carnot heat engine on the difference in the reservoir temperatures. Explain why this is so.

A box is separated by a partition into two parts of equal volume. The left side of the box contains 500 molecules of nitrogen gas; the right side contains 100 molecules of oxygen gas. The two gases are at the same temperature. The partition is punctured, and equilibrium is eventually attained. Assume that the volume of the box is large enough for each gas to undergo a free expansion and not change temperature. (a) On average, how many molecules of each type will there be in either half of the box? (b) What is the change in entropy of the system when the partition is punctured? (c) What is the probability that the molecules will be found in the same distribution as they were before the partition was punctured-that is, 500 nitrogen molecules in the left half and 100 oxygen molecules in the right half?

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