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Three moles of an ideal gas undergo a reversible isothermal compression at 20.0\(^\circ\)C. During this compression, 1850 J of work is done on the gas. What is the change of entropy of the gas?

Short Answer

Expert verified
The change in entropy is approximately 6.31 J/K.

Step by step solution

01

Understand the problem

We need to find the change in entropy of an ideal gas when it undergoes a reversible isothermal compression. Given are the number of moles (3 moles), temperature (20.0°C), and work done on the gas (1850 J).
02

Convert Temperature to Kelvin

To use the temperature in thermodynamic equations, we need to convert it from Celsius to Kelvin. The conversion is given by: \[ T(K) = T(°C) + 273.15 \]Thus,\[ T(K) = 20.0 + 273.15 = 293.15\, K. \]
03

Use the First Law of Thermodynamics

For an isothermal process, the change in internal energy is zero for an ideal gas (\( \Delta U = 0 \)). According to the first law of thermodynamics:\[ \Delta U = Q - W, \]where \( \Delta U = 0 \), \( Q \) is heat absorbed, and \( W \) is work done on the gas. This implies:\[ Q = W, \]Therefore, \( Q = 1850\, J \).
04

Compute the change in entropy

The change in entropy (\( \Delta S \)) for a reversible isothermal process is given by:\[ \Delta S = \frac{Q}{T}, \]Substituting the known values:\[ \Delta S = \frac{1850}{293.15} \approx 6.31\, J\/K. \]

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Key Concepts

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

Ideal Gas
In our scenario, we're dealing with an ideal gas, which is a theoretical concept often used in physics and chemistry. An ideal gas is made up of many randomly moving particles that are far enough apart so they do not exert any force on each other, except during collisions. These collisions, however, are perfectly elastic. This means when particles bump into each other or the walls of a container, they don't lose any energy.

There are several important assumptions about ideal gases:
  • The gas consists of a large number of small particles separated by distances far greater than their size.
  • These particles are in constant, random motion.
  • All collisions between particles, or with the walls, are perfectly elastic (no energy loss).
  • There are no attractive or repulsive forces between the particles except during collisions.
  • The gas adheres to the ideal gas law, PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature in Kelvin.
This simple model helps us predict how gases will behave under different conditions, and it's extremely helpful when solving problems like the one in this exercise.
Isothermal Process
An isothermal process is a type of thermodynamic process in which the temperature of the system remains constant. In the context of our problem, the ideal gas undergoes an isothermal compression. Here's what makes the process distinct:
  • The temperature (T) remains constant throughout the process. Since we know temperature dictates how much kinetic energy particles have, maintaining a constant temperature implies energy input/output must balance any work being done on/by the system.
  • This balance of energy in an isothermal process typically means heat is exchanged between the system and its surroundings. This exchange ensures that the internal energy stays unchanged, despite any work done on the gas.
  • For ideal gases, this means the internal energy change (ΔU) is zero because internal energy is dependent solely on temperature in an ideal gas.
For the problem at hand, when the ideal gas is compressed, work is done on it, but because temperature is constant (thanks to energy exchange), it remains an isothermal compression, showing the elegance of thermodynamic processes.
First Law of Thermodynamics
The first law of thermodynamics is a crucial principle in understanding processes like isothermal compression. This law, often described as the law of energy conservation, states that energy can neither be created nor destroyed; it can only be transformed from one form to another. Mathematically, it is expressed as:
  • \[\Delta U = Q - W\]where \( \Delta U \) is the change in internal energy of the system, \( Q \) is the heat added to the system, and \( W \) is the work done by the system.
  • In an isothermal process for an ideal gas, \( \Delta U = 0 \) because the internal energy depends solely on the temperature. Since the temperature is constant, the change in internal energy is zero.
  • This simplifies the first law to \( Q = W \). This equation tells us that, in an isothermal process, the heat added to the system is equal to the work done on the system.
In the original problem, applying this principle allowed us to compute how much heat was exchanged (or absorbed) during the compression. Understanding this concept helps clarify how energy moves and changes form during different thermodynamic processes.

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Most popular questions from this chapter

You make tea with 0.250 kg of 85.0\(^\circ\)C water and let it cool to room temperature (20.0\(^\circ\)C). (a) Calculate the entropy change of the water while it cools. (b) The cooling process is essentially isothermal for the air in your kitchen. Calculate the change in entropy of the air while the tea cools, assuming that all of the heat lost by the water goes into the air. What is the total entropy change of the system tea + air?

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?

A typical coal-fired power plant generates 1000 MW of usable power at an overall thermal efficiency of 40%. (a) What is the rate of heat input to the plant? (b) The plant burns anthracite coal, which has a heat of combustion of 2.65 \(\times\) 10\(^7\) J/kg. How much coal does the plant use per day, if it operates continuously? (c) At what rate is heat ejected into the cool reservoir, which is the nearby river? (d) The river is at 18.0\(^\circ\)C before it reaches the power plant and 18.5\(^\circ\)C after it has received the plant's waste heat. Calculate the river's flow rate, in cubic meters per second. (e) By how much does the river's entropy increase each second?

A gasoline engine takes in 1.61 \(\times\) 10\(^4\) J of heat and delivers 3700 J of work per cycle. The heat is obtained by burning gasoline with a heat of combustion of 4.60 \(\times\) 10\(^4\) J/g. (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?

Compare the entropy change of the warmer water to that of the colder water during one cycle of the heat engine, assuming an ideal Carnot cycle. (a) The entropy does not change during one cycle in either case. (b) The entropy of both increases, but the entropy of the colder water increases by more because its initial temperature is lower. (c) The entropy of the warmer water decreases by more than the entropy of the colder water increases, because some of the heat removed from the warmer water goes to the work done by the engine. (d) The entropy of the warmer water decreases by the same amount that the entropy of the colder water increases.

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