Problem 27 - Pearson Physics

Question

Calculate How much heat must be added to a gas that does \(10 \mathrm{~J}\) of work in a constant-temperature (isothermal) process?

Step-by-Step Solution

Answer

The heat added is \( 10 \mathrm{~J} \).

Step 1: Understand the Process

In an isothermal process, the temperature of the gas remains constant. According to the first law of thermodynamics, the change in internal energy for an isothermal process is zero, because the internal energy of an ideal gas depends only on its temperature.

Step 2: Identify Relevant Formula

The first law of thermodynamics can be written as: \[ \Delta U = Q - W \] where \( \Delta U \) is the change in internal energy, \( Q \) is the heat added to the system, and \( W \) is the work done by the system. Since \( \Delta U = 0 \) for an isothermal process, the equation simplifies to \( Q = W \).

Step 3: Substitute Known Values

You are given that the work done \( W = 10 \mathrm{~J} \). According to the simplified equation from Step 2, \( Q = W \). Therefore, substitute the value of \( W \) into the equation: \( Q = 10 \mathrm{~J} \).

Step 4: Interpret the Result

The result tells us the amount of heat that must be added to the gas to keep the temperature constant while it does \(10 \mathrm{~J}\) of work. This implies that the heat added compensates exactly for the work done, maintaining the internal energy constant.

Key Concepts

Isothermal Process

An isothermal process is a fascinating concept in thermodynamics where the temperature of the system remains constant. But how does that happen whilst the process evolves? Think of it as a balancing act where the energy added as heat (Q) perfectly matches the energy lost as work (W). This means there's no change in internal energy since it depends solely on temperature for an ideal gas.
In the context of the first law of thermodynamics, the equation is:

• \[ \Delta U = Q - W \]

Here, \( \Delta U \) represents the change in internal energy, \( Q \) is the heat added, and \( W \) is the work done by the system. Since \( \Delta U = 0 \) during an isothermal process, every joule of work done is offset by a joule of heat added, preserving the temperature of the gas. This interplay makes isothermal processes essential in understanding the behavior of gases under controlled conditions.

Internal Energy

Internal energy is a core concept in understanding thermodynamic processes. For a given system, it represents the sum of all microscopic forms of energy within it, such as kinetic and potential energy of the molecules. The fascinating part about an isothermal process is that despite the heat and work interactions, the internal energy remains unchanged.
Let's delve deeper: in an ideal gas, internal energy is directly linked to its temperature, and because the temperature stays constant in an isothermal process, the internal energy doesn't change. In mathematical terms, this is expressed as:

• \( \Delta U = 0 \)

This zero change means any heat (Q) added to the system is completely offset by the work (W) done, perfectly maintaining the gas's initial energy state. Understanding this delicate balance helps in grasping the intricate mechanics of the first law of thermodynamics.

Work and Heat

In the context of thermodynamics, work and heat are two forms of energy transfer. But what sets them apart, and how do they interact? Let's break it down.

• **Work**: This is energy transferred by a system's boundary movement. In our exercise, when the gas does work, it's performing energy transfer to its surroundings, which is given as 10 J.
• **Heat**: Contrarily, this is energy transferred because of temperature difference—not that the system's temperature changes in isothermal processes. Here, heat supplied compensates the work done, to maintain the system's temperature.

For an isothermal process:

• \( Q = W \)

This equation implies the heat added to the system equals the work done by it. In our example, adding 10 J of heat balances the 10 J of work done, ensuring no overall change in internal energy. This balance illustrates the beautiful simplicity of energy conservation under isothermal conditions.