Problem 25 - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

The change in entropy is approximately 6.31 J/K.

Step 1: 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).

Step 2: 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,\[ T(K) = 20.0 + 273.15 = 293.15\, K. \]

Step 3: 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, \( Q = 1850\, J \).

Step 4: 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:\[ \Delta S = \frac{1850}{293.15} \approx 6.31\, J\/K. \]

Key Concepts

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.