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Physics

Chapter 15: 1-4

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Thermodynamics

Introduction

A theory is the more impressive the greater the simplicity of its premises, the more different kinds of things it relates, and the more extended its area of applicability. Therefore the deep impression that classical thermodynamics made upon me. It is the only physical theory of universal content which I am convinced will never be overthrown, within the framework of applicability of its basic concepts.

— Albert Einstein, Autobiographical Notes

Outline

The Laws of Thermodynamics

We've been studying the conditions under which gas characteristics change. But a gas is in many ways a model for any system which interacts as a whole with its environment. We've seen how mechanical energy, the sum of the kinetic and potential energy of a system, is conserved under certain circumstances. Now we move to an even more general case, where we look at not only mechanical energy or work, but also heat energy, energy flows that depend on changes in temperature.

Systems

Before we can talk about thermodynamics, we need to discuss the conditions under which we measure energy flow. Practical and theoretical applications focus on the concept of systems, sets of objects within defined boundaries. Energy flow is measured across the boundary between the system and the environment (everything not inside the system boundaries).

First Law of Thermodynamics

According to the first law of thermodynamics, any change in internal energy is accomplished either by heat flow in or out of the system, or by work done on or by the system. We can symbolize this with

Δ KE   + Δ PE   + Δ U   =   Q     W

where

Example: A system consists of 3.0 kg of water at 80°C. 25J of work is done on the system by stirring the water with a paddle wheel, and 63J of heat is removed. What is the internal energy of the system?

Identify the heat source/sink 63 J of heat are removed: Q is NEGATIVE
Identify work done by/on system 25 J of work done on the system: W is NEGATIVE
Set up the equation and substitute δU = Q - W = - 63J - (- 25J) = -38 J

In this case, we have a closed system: both heat and work energy flow across the boundaries, but (apparently) no matter. In an isolated system, with no flow across the boundaries, δU must be zero in order to conserve energy, and Q = W. There are other intersting situations...

Applications

Consider a gas in a container, subject to the gas law PV = nRT. We can put it through a number of processes, with varying conditions.

Let's envision a piston of circular surface area A, subject to force F which pushes the piston down a distance d.

In the isobaric process, the amount of work done is derived from the definitions for work and pressure: W = F * d. But from the relationship Pressure P = F/A, we can substitute W = PA * d. Now A*d is just the volume through which the force has acted. Since P is held constant, W = P * δV. There is work done on the system (if V decreases) or by the system (if V increases). There may also be a heat flow of Q, where Q is not zero. So for an isobaric process, δU = Q - W = Q - P * δV.

In the isochoric or isometric process, δV is zero, so W = P*δV is zero: no work is done. Any isochoric process occurs if and only if there is a heat exchange with the outside universe. δU = Q .

Practice with the Concepts

Which gives the greater improvement in the efficiency of an ideal (Carnot) engine: a decrease in the low-temperature reservoir, or the same amount of increase in the high-temperature reservoir?

Discussion Points