Problem 31 - College Physics

Question

Three moles of an ideal gas undergo a reversible isother- mal compression at \(20.0^{\circ} \mathrm{C}\) . During this compression, 1850 \(\mathrm{J}\) of work is done on the gas. What is the change in 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 have an ideal gas undergoing a reversible isothermal process at a constant temperature of \(20.0^{\circ}\, \mathrm{C}\). A work of 1850 J is done on the gas, and we need to find the change in entropy.

Step 2: Recall the Isothermal Process Formula

For an isothermal process, the change in internal energy of an ideal gas is zero because temperature is constant. Thus, the work done on the gas \(W\) is equal to the heat exchanged \(Q\). Hence, \(Q = W = 1850 \text{ J}\).

Step 3: Use Entropy Change Formula

The change in entropy for a reversible process is given by \(\Delta S = \frac{Q}{T}\) where \(Q\) is the heat exchange and \(T\) is the absolute temperature in Kelvin.

Step 4: Convert Temperature to Kelvin

Convert the temperature from Celsius to Kelvin: \(T = 20^{\circ}\, \mathrm{C} + 273.15 = 293.15\, \mathrm{K}\).

Step 5: Calculate the Change in Entropy

Substitute \(Q = 1850 \text{ J}\) and \(T = 293.15\, \mathrm{K}\) into the formula: \[\Delta S = \frac{1850}{293.15}\.\] Calculate this to find \(\Delta S\).

Step 6: Compute the Final Answer

Perform the calculation: \(\Delta S = \frac{1850}{293.15} \approx 6.31 \text{ J/K}\).

Key Concepts

Isothermal Process

An isothermal process occurs when a system, such as an ideal gas, undergoes a change where the temperature remains constant. Since temperature does not change, the internal energy of the gas stays the same. This type of process is often carried out slowly enough to keep the system in thermal equilibrium with its surroundings.
In an isothermal process involving an ideal gas, any work done on or by the system is balanced by heat transfer. This means that whatever energy is put into the system as work must come out as heat, or vice versa. For example, if we compress a gas isothermally, the energy used to do this work converts directly into heat that leaves the gas, maintaining a constant temperature throughout.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in thermodynamics that relates the pressure, volume, and temperature of an ideal gas. It is expressed mathematically as: \[ PV = nRT \]where:

• \( P \) is the pressure of the gas,
• \( V \) is the volume,
• \( n \) is the number of moles,
• \( R \) is the ideal gas constant, and
• \( T \) is the temperature in Kelvin.

This equation helps to predict how a gas will behave under different conditions. It assumes the gas in question is "ideal," meaning it perfectly follows the laws of kinetic theory, and the molecules do not interact with each other in terms of attraction or repulsion. In real-world applications, this law serves as a valuable approximation that simplifies complex calculations, especially for processes like isothermal expansions and compressions.

Reversible Process

A reversible process is a theoretical concept in thermodynamics where a system changes state in such a way that the process can be exactly reversed without leaving any net change in the system or the surrounding environment. This means that both the system and its surroundings return to their original states.

• Reversible processes occur infinitely slowly so that the system always remains in equilibrium.
• They are represented as a perfectly smooth and continuous transition in the thermodynamic path.
• For entropy changes, reversible processes allow precise calculations using formulas like \( \Delta S = \frac{Q}{T} \).

While true reversible processes are idealizations, they set the standard for maximum efficiency and help us understand the thermodynamic limits of real-world processes.

Work and Heat in Thermodynamics

Understanding work and heat is crucial in studying thermodynamics as they represent two modes of energy transfer. In the context of an isothermal and reversible process involving an ideal gas:

• **Work**: When work is done on or by a gas, it involves a force applied over a distance. In our exercise, 1850 J of work is done on the gas during compression. This work changes either the volume or the pressure of the gas, directly affecting its energy state.
• **Heat**: Representing energy transfer due to temperature difference, heat will flow into or out of the gas to maintain the constant temperature in an isothermal process. For our system, the heat exchange equals the work done because there is no change in internal energy (since temperature is constant).

Understanding how work and heat relate through isothermal actions allows us to grasp more complex thermodynamic concepts and solves practical problems such as entropy change in a straightforward manner.