Problem 12 - University Physics with Modern Physics

Question

A gas in a cylinder is held at a constant pressure of \(1.80 \times 10^{5} \mathrm{Pa}\) and is cooled and compressed from 1.70 \(\mathrm{m}^{3}\) to 1.20 \(\mathrm{m}^{3} .\) The intermal energy of the gas decreases by \(1.40 \times 10^{5} \mathrm{J}\) . (a) Find the work done by the gas. (b) Find the absolute value \(Q\) of the heat flow into or out of the gas, and state the direction of the heat flow. (c) Does it matter whether the gas is ideal? Why or why not?

Step-by-Step Solution

Answer

The work done by the gas is \(-9.0 \times 10^4 \mathrm{J}\), heat flow is \(-2.30 \times 10^5 \mathrm{J}\) (out of the gas), and gas type doesn't affect the solution.

Step 1: Understand the Question and Given Data

The exercise is about a gas undergoing a process where it is compressed and its internal energy changes. We need to determine:1. The work done by the gas.2. The heat flow into or out of the gas and its direction.3. If the nature of the gas (ideal vs non-ideal) affects this problem.Given data:- Constant pressure: \(P = 1.80 \times 10^{5} \mathrm{Pa}\)- Initial volume: \(V_1 = 1.70 \, \mathrm{m}^{3}\)- Final volume: \(V_2 = 1.20 \, \mathrm{m}^{3}\)- Change in internal energy: \(\Delta U = -1.40 \times 10^{5} \, \mathrm{J}\) (a decrease implies it is negative).

Step 2: Calculate the Work Done by the Gas

The work done by a gas at constant pressure is given by the formula:\[ W = P \cdot (V_2 - V_1) \]Substitute the given values:\[ W = 1.80 \times 10^5 \, \text{Pa} \times (1.20 \, \text{m}^3 - 1.70 \, \text{m}^3) \]Calculate:\[ W = 1.80 \times 10^5 \, \text{Pa} \times (-0.50 \, \text{m}^3) \]\[ W = -9.0 \times 10^4 \, \text{J} \]The negative sign indicates work is done on the gas.

Step 3: Determine the Heat Flow and Direction

We apply the first law of thermodynamics which is given by:\[ \Delta U = Q - W \]Where:- \( \Delta U \) is the change in internal energy.- \( Q \) is the heat added to the system.- \( W \) is the work done by the system.Rearrange the equation to solve for \( Q \):\[ Q = \Delta U + W \]Using \( \Delta U = -1.40 \times 10^{5} \mathrm{J} \) and \( W = -9.0 \times 10^{4} \, \mathrm{J} \):\[ Q = (-1.40 \times 10^{5} ) + (-9.0 \times 10^4) \]\[ Q = -2.30 \times 10^5 \, \mathrm{J} \]The negative sign indicates that the heat flows out of the gas.

Step 4: Does it Matter if the Gas is Ideal?

In this problem, it does not matter whether the gas is ideal or not because we are using the first law of thermodynamics which applies to all types of gases under constant pressure and changes in volume conditions. The calculation of work and heat depends on pressure, volume, and internal energy change, all of which apply to both ideal and real gases.

Key Concepts

First Law of Thermodynamics

The First Law of Thermodynamics is a fundamental principle in physics that provides the framework for understanding energy transformations within a system. It states that energy cannot be created or destroyed, only transformed from one form to another. In equation form, the first law is expressed as: \( \Delta U = Q - W \), where:

• \( \Delta U \) is the change in internal energy of the system.
• \( Q \) represents the heat added to the system.
• \( W \) is the work done by the system.

In the given problem, the gas is compressed, which affects both its internal energy and transfers energy in the form of heat and work. The equation helps us understand how much heat flows in or out of the gas, based on its internal energy change and the work performed.

Ideal Gas

The concept of an ideal gas is a theoretical model that simplifies the behavior of gases by assuming no intermolecular forces and that the gas molecules occupy no volume themselves. While no real gas perfectly fits this description, ideal gas laws are incredibly useful for calculations and understanding basic thermodynamic processes.
For our exercise, whether the gas behaves as an ideal gas or not does not impact our calculations under constant pressure and volume change conditions. We are primarily concerned with changes in state quantities like pressure, volume, and internal energy, which are applicable to both ideal and real gases.
Although ideal gas assumptions simplify models, it is crucial to realize that they only approximate real-world behavior under many conditions. This understanding allows physicists and engineers to make accurate predictions in various practical situations.

Internal Energy Change

Internal energy change refers to the difference in a system's total energy due to changes in state or conditions such as pressure, volume, or temperature. It represents the total kinetic and potential energy of the molecules within the gas.
In the problem, the internal energy decreases by \(-1.40 \times 10^{5}\, \text{J}\). This negative change signals that the gas has lost energy. This reduction can occur from work being done on the gas and/or heat being removed from the system.
The internal energy change plays a pivotal role in determining whether heat must be added or removed from the gas to maintain thermodynamic equilibrium or to perform a desired task like a power cycle.

Work Done by Gas

The work done by a gas when it expands or compresses can be calculated using the relation \( W = P \, \Delta V \), where \( P \) is the constant pressure and \( \Delta V \) is the change in volume. In this exercise, the gas is compressed, resulting in a negative change in volume.

Calculating work done:

• Initial volume \( V_1 = 1.70 \, \text{m}^3 \)
• Final volume \( V_2 = 1.20 \, \text{m}^3 \)
• This gives \( \Delta V = V_2 - V_1 = -0.50 \, \text{m}^3 \)
• With pressure \( P = 1.80 \times 10^{5} \text{Pa} \), the work done is \( W = 1.80 \times 10^{5} \text{Pa} \times (-0.50 \, \text{m}^3) = -9.0 \times 10^4 \, \text{J} \)

The negative sign indicates the work is done on the gas, reflecting that energy is transferred to it due to compression. Understanding work done by or on a gas is crucial for processes such as engines and refrigeration cycles.