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Chapter 15: Problem 10

A system gains 2780 J of heat at a constant pressure of \(1.26 \times\) \(10^{5} \mathrm{Pa},\) and its internal energy increases by \(3990 \mathrm{J} .\) What is the change in the volume of the system, and is it an increase or a decrease?

Short Answer

Expert verified
Change in volume: \(-0.0096 \text{ m}^3\); the volume increases.

Step by step solution

01

Understanding the Problem

We are given a system that gains heat and experiences a change in internal energy. We need to find the change in volume and determine if it's an increase or decrease.
02

Applying the First Law of Thermodynamics

The First Law of Thermodynamics states: \( \Delta U = Q - W \), 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. We have \( \Delta U = 3990 \text{ J} \) and \( Q = 2780 \text{ J} \).
03

Solving for Work Done

Rearrange the equation to solve for work done by the system: \( W = Q - \Delta U \). Substitute \( Q = 2780 \text{ J} \) and \( \Delta U = 3990 \text{ J} \) into the equation: \( W = 2780 \text{ J} - 3990 \text{ J} = -1210 \text{ J} \).
04

Understanding Work Done in Terms of Volume Change

Work done by the system at constant pressure is given by \( W = P \Delta V \), where \( P \) is pressure and \( \Delta V \) is the change in volume. Since \( W = -1210 \text{ J} \) and \( P = 1.26 \times 10^5 \text{ Pa} \), we see that work done is negative, indicating volume increase.
05

Calculating Change in Volume

Rearrange \( W = P \Delta V \) to solve for \( \Delta V \): \( \Delta V = \frac{W}{P} = \frac{-1210 \text{ J}}{1.26 \times 10^5 \text{ Pa}} = -0.0096 \text{ m}^3 \).
06

Interpreting the Sign of Change in Volume

Since \( \Delta V = -0.0096 \text{ m}^3 \) is negative in our equation setup, this actually indicates an increase in volume as the system did work (expanding).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Internal Energy Change
Internal energy change is a fundamental concept in thermodynamics, and it refers to the amount of energy stored within a system. When heat is added to or removed from a system, or work is done by or on the system, its internal energy changes. In our scenario:
  • The system's internal energy increased by 3990 Joules.
  • This suggests that the energy within the system grew due to a combination of heat and work interactions.
Understanding internal energy is crucial because it helps us quantify and predict how a system behaves under different conditions, like when energy is poured into or drawn away from it. The tricky part is to account for all forms of energy that contribute to this change.
Heat Transfer
Heat transfer is the process of energy moving from a warmer object to a cooler one. In the context of our exercise, 2780 J of heat was transferred into the system at a constant pressure. This energy transfer method plays a vital role in changing the system's state:
  • Heat, denoted by Q, is added to the system, increasing its thermal energy.
  • The addition of heat often results in temperature change, phase change, or as in this case, it changes the internal energy of the system.
Heat transfer, as seen here, is a driving factor in altering a system's internal energy and is a foundational aspect of the First Law of Thermodynamics.
Work-Energy Principle
The work-energy principle in thermodynamics is closely woven with the concept of work done by or on a system. In simple terms, when a system does work, it transfers energy to its surroundings. Given the formula: \[ W = P \Delta V \]We see how work is related to changes in volume. The negative sign of work done \(-1210 \text{ J}\) signals that the system's volume has increased:
  • The work-energy principle dictates that if a system expands, it must do work against its surroundings.
  • This work done results in an increase in the volume of the system鈥攊ndicated by the negative work calculated.
Understanding this principle helps us determine whether a system is doing work, like expanding, or having work done onto it, like compression, depending on the signs of work and pressure-volume changes.

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Most popular questions from this chapter

A diesel engine does not use spark plugs to ignite the fuel and air in the cylinders. Instead, the temperature required to ignite the fuel occurs because the pistons compress the air in the cylinders. Suppose that air at an initial temperature of \(21^{\circ} \mathrm{C}\) is compressed adiabatically to a temperature of \(688^{\circ} \mathrm{C} .\) Assume the air to be an ideal gas for which \(\gamma=\frac{7}{5} .\) Find the compression ratio, which is the ratio of the initial volume to the final volume.

You and your team are tasked with evaluating the parameters of a high- performance engine. The engine compresses the fuel-air mixture in the cylinder to make it ignite rather than igniting it with a spark plug (this is how diesel engines operate). You are given the compression ratio for the cylinders in the engine, which is the ratio of the initial volume of the cylinder (before the piston compresses the gas) to the final volume of the cylinder (after the gas is compressed): \(V_{i}: V_{\mathrm{f}}\). Assume that the fuel- gas mixture enters the cylinder at a temperature of \(22.0^{\circ} \mathrm{C}\), and that the gas behaves like an ideal gas with \(\gamma=7 / 5 .\) (a) If the compression ratio is \(15.4: 1,\) what is the final temperature of the gas if the compression is adiabatic? (b) With everything else the same as in (a), what is the final temperature of the gas if the compression ratio is increased to \(17.0: 1 ?\) (c) Everything else being the same, for which compression ratio do you think the engine runs more efficiently? Give a qualitative argument for your answer.

Argon is a monatomic gas whose atomic mass is 39.9 u. The temperature of eight grams of argon is raised by \(75 \mathrm{K}\) under conditions of constant pressure. Assuming that argon behaves as an ideal gas, how much heat is required?

A monatomic ideal gas in a rigid container is heated from \(217 \mathrm{K}\) to \(279 \mathrm{K}\) by adding \(8500 \mathrm{J}\) of heat. How many moles of gas are there in the container?

A Carnot air conditioner maintains the temperature in a house at \(297 \mathrm{K}\) on a day when the temperature outside is \(311 \mathrm{K}\). What is the coefficient of performance of the air conditioner?
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