In Figure P 20.40 , the change in internal energy of a gas...

In Figure $\mathrm{P} 20.40$ , the change in internal energy of a gas that is taken from $A$ to $C$ is $+800 \mathrm{J}$ . The work done on the gas along path $A B C$ is $-500 \mathrm{J}$ . (a) How much energy must be added to the system by heat as it goes from $A$ through $B$ to $C ?$ (b) If the pressure at point $A$ is five times that of point $C,$ what is the work done on the system in going from $C$ to $D P$ (c) What is the energy exchanged with the surroundings by heat as the cycle goes from $C$ to $A$ along the green path? (d) If the change in internal energy in going from point $D$ to point $A$ is $+500 \mathrm{J}$ , how much energy must be added to the system by heat as it goes from point $C$ to point $D ?$

Key Concepts

First Law of Thermodynamics

The First Law states that energy cannot be created or destroyed, only transformed. In thermodynamics, this law is expressed as the change in internal energy of a system being equal to the heat added to the system plus the work done on it. This principle is fundamental because it underlies all energy balance calculations in thermodynamic processes.

Sign Conventions in Thermodynamics

Understanding sign conventions is crucial in thermodynamics. Typically, work done by the system is considered positive, while work done on the system is negative. Similarly, heat added to the system is positive, and heat removed is negative. These conventions are essential for accurately applying the First Law of Thermodynamics across different processes.

Work in Thermodynamic Processes

Work in thermodynamics refers to the energy transfer that occurs when a force moves a boundary of the system, such as in compression or expansion of gases. The amount of work done is calculated by integrating the pressure with respect to the change in volume in the case of quasi-static processes, and it plays a key role in the energy balance described by the First Law.

Heat Transfer in Thermodynamic Processes

Heat transfer involves the movement of thermal energy between systems or within different parts of a system due to a temperature difference. It is one of the primary modes of energy exchange in thermodynamic processes and, when combined with work, determines the overall energy change of a system.

Thermodynamic Cycles

A thermodynamic cycle is a series of processes that return a system to its initial state. Despite the net change in internal energy being zero over one complete cycle, heat and work exchanges still occur in each individual process. Analyzing cycles is important for understanding engines and refrigerators, where the performance depends on managing both work and heat transfer across various stages.

Transcript

00:11 In this example, i'm going to be looking at the expansion and compression cycle of a gas.

00:17 We're going to be looking at the energy of the gas, the heat added or taken away from the gas, and the work done to it or by it.

00:25 So let me show you what our pv diagram looks like for one second, or sorry, one cycle, volume, pressure.

00:34 We start up here at point a at some initial pressure and volume.

00:40 We expand to point b here.

00:46 Then we're going to cool down.

00:49 My pressure is going to drop until we get to point c here.

00:54 Then we're going to compress over here to point d.

00:59 And finally, come back to our original state at point a.

01:06 And we're going to be given some information about the system and asked to find some other information.

01:11 First problem tells us that as we go from point a to c, so from a to c along path a, b, c, so a to b to c, my change in internal energy equals 800 joules.

01:32 And my work done, delta w, as the gas expands, is negative 500 joules.

01:42 And i want to know how much heat is added as we go from point a here to point c here.

01:53 I know that the change in internal temperature for a gas equals the heat in plus the work on.

02:06 My numbers, 800 joules equals q in minus 500 joules.

02:16 It's negative because the work does work.

02:19 It's not being worked on.

02:21 So the heat added to the system is going to be 1 ,300 joules.

02:27 The second question tells us that my pressure, initial pressure, equals five times my final pressure.

02:40 So initial pressure we have up here.

02:44 That's the pressure from point a to b.

02:47 Final pressure is down here, the pressure of both points c and d.

02:53 And we want to find the work done on the gas to compress it back from point c to d.

03:03 Right, so we know work equals p delta v, or p is the pressure.

03:14 At that point, delta v is the final volume minus the initial.

03:19 Looking at the work done from point a to point b of work equals i'll call this pa times vb minus va and we already said that equals negative 500 joules right and for my work going from i'll call this pc pc v final minus v initial.

04:00 That equals, we don't know.

04:03 I can make a couple of substitutions.

04:05 I can use the relationship between initial pressure and final pressure to rewrite this equation as w equals 1 over 5 p .a.

04:19 And i'll use the fact that the volume at point d equals a volume at point a.

04:25 Write va minus the same thing with b and c...

Please give Ace some feedback