Question

The work done to compress a gas is \(47 \mathrm{~J}\). As a result, \(93 \mathrm{~J}\) of heat is given off to the surroundings. Calculate the change in internal energy of the gas.

Short answer

The change in internal energy of the gas is \(-46 \mathrm{~J}\).

Step-by-step solution

Identify Known Values

We are given two important values in the problem: the work done on the gas is given as \( W = 47 \mathrm{~J} \) and the heat given off to the surroundings is \( Q = -93 \mathrm{~J} \) (since heat is leaving the system).

Apply the First Law of Thermodynamics

The first law of thermodynamics is given by the formula \( \Delta U = Q + W \), where \( \Delta U \) is the change in internal energy, \( Q \) is the heat exchange, and \( W \) is the work done on or by the system. In this context, heat leaves the system, so \( Q \) is negative.

Substitute Known Values into the Equation

Substitute \( Q = -93 \mathrm{~J} \) and \( W = 47 \mathrm{~J} \) into the first law equation: \[ \Delta U = (-93 \mathrm{~J}) + 47 \mathrm{~J} \].

Calculate the Change in Internal Energy

Perform the arithmetic: \[ \Delta U = -93 + 47 = -46 \mathrm{~J} \]. This result means the internal energy of the gas decreases by 46 Joules.

Key concepts

First Law of Thermodynamics

The First Law of Thermodynamics is a fundamental principle that links the concepts of energy and matter. It states that energy cannot be created or destroyed in an isolated system, it can only be transformed from one form to another. In other words, the total energy of an isolated system remains constant. This law helps us understand how energy flows within different systems in the universe.

For practical purposes, this law is expressed using the formula:

• \( \Delta U = Q + W \)

Here, \( \Delta U \) represents the change in internal energy, \( Q \) is the heat added to the system, and \( W \) is the work done on or by the system.

Knowing this equation allows us to calculate the change in internal energy when we know the heat and work values. This becomes critical in thermodynamics problems, where understanding energy transformations is key.

Internal Energy

Internal energy is the total energy contained within a system due to the random motions of its molecules. It encompasses both potential and kinetic energy of the molecules. This energy is an intrinsic property of a system, depending on the state variables such as temperature and volume.

When a system undergoes an energy transfer—like a gas being compressed—its internal energy is affected. Any increase in temperature typically corresponds to an increase in internal energy. Conversely, if a gas cools down, its internal energy decreases. Thus, the change in internal energy is crucial for understanding thermodynamic processes.

In the context of our problem, internal energy decreases since energy leaves the system in the form of heat.

Heat Exchange

Heat exchange refers to the transfer of thermal energy between a system and its surroundings. This can happen in several ways:

• Conduction: Transfer through direct contact.
• Convection: Transfer through fluid motion.
• Radiation: Transfer through electromagnetic waves.

When heat is added to a system, it can increase the system's temperature or do work. Similarly, when a system loses heat, it may perform less work or lower its internal energy.

In our exercise, the gas releases heat to the surroundings, quantified as \( Q = -93 \mathrm{~J} \). Since heat is lost, it's labeled as negative, indicating a reduction in the system's energy.

Work Done on Gas

Work done on or by a gas is a measure of energy transfer resulting from volume changes in the gaseous system. In thermodynamics, work is involved whenever a gas expands or compresses.

• If the gas is compressed, work is done on it.
• If the gas expands, it does work on its surroundings.

The formula for calculating work, when the volume of gas changes at constant pressure, is:

• \( W = P \Delta V \)

For the problem at hand, \( W = 47 \mathrm{~J} \) indicates that work is done on the gas, increasing its energy. However, since heat is simultaneously lost, the internal energy ultimately decreases.

This interplay between heat and work explains why understanding both is essential for solving thermodynamic equations effectively.