In Figure P 20.36 , the change in internal energy of a gas...

In Figure $\mathrm{P} 20.36$ , 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 $D ?$ (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 $+500$ J, how much energy must be added to the system by heat as it goes from point $C$ to point $D ?$

Key Concepts

Cyclic Processes

A cyclic process is one in which the system undergoes a series of changes and returns to its initial state. In such a process, the net change in internal energy is zero over the entire cycle, meaning the total work done by the system is exactly balanced by the total heat exchanged. This concept is fundamental in the analysis of engines and refrigerators where repeated cycles are essential.

Pressure-Volume Work

Pressure-volume work occurs when a gas expands or compresses, causing a change in volume under a given pressure. This type of work is directly related to changes in the system's energy and can be calculated as the integral of pressure with respect to volume change, often under specific process constraints like constant pressure or varying pressure scenarios.

Work Done in Thermodynamic Processes

Work in thermodynamics involves energy transfer that results from a force acting through a displacement, typically seen as pressure-volume work in gases. The sign conventions of work—whether work is done by the system or on the system—are crucial for correctly applying the energy balance provided by the First Law of Thermodynamics.

Heat Transfer

In thermodynamic processes, heat transfer refers to the energy exchanged between a system and its surroundings due to a temperature difference. It plays a critical role in changing the internal energy of a system and is one of the primary ways energy is added to or removed from a system.

Internal Energy as a State Function

Internal energy is a property of a system that depends only on its current state and not on the path by which the system arrived at that state. This means changes in internal energy can be calculated solely from the initial and final states, making it an essential concept in analyzing and comparing different thermodynamic paths between the same end points.

First Law of Thermodynamics

This principle of energy conservation in thermodynamics states that the change in the internal energy of a system is equal to the heat added to the system minus the work done by the system. It serves as the fundamental energy balance equation for all thermodynamic processes, ensuring that energy is neither created nor destroyed but simply transformed from one form to another.

Transcript

00:01 Consider a gas as described by figure p20 .36, or this pv diagram up to the left.

00:06 And we also know that the change in internal energy between points a and c is given to us as 800 joules.

00:17 And the work between points a, b, and c is given to us as negative 500 joules.

00:25 So first, let's try and figure out what heat must be added to the system as it goes along the red path from a to b to c.

00:35 That is, what is q sub a, b, c? and we're going to just take a look at the first law firmer dynamics to kick off, where in the delta, the change in internal energy is given by just the heat plus the work.

00:51 And it's worth noting that while the heat and the work are both path -dependent quantities, the change in internal energy is not.

01:03 It is a path independent value.

01:05 That means it doesn't matter if i take the red path from a to b to c or the blue path from a to c.

01:11 The change in internal energy between points a and c is the same regardless.

01:16 So i can go ahead and sub in our given 800 joules for the internal energy.

01:22 And this is q sub a, b, c, and then just our given work from a .b .c.

01:30 This is a very simple calculation to get you that q sub a, b, c is just 1 ,300 joules.

01:40 Now, what else can we figure out? if it's given to us that the pressure at points a and b is five times the pressure at point c and d, that is, let's say, pab equals 5 p cd, then can we figure out the work required to go from point c to point d? and it's worth pausing for a sec to take a look at each of the steps of this process we have going on.

02:13 Namely, as we go from a to b, we have a constant pressure and a changing volume, which you might recognize as an isobaric process, which is, as its work described by the function, negative pressure times the final volume minus the initial volume.

02:35 And on the other hand, going from b to c and from d to a, you might notice that we have changing pressures but constant volumes, which tells you that it's an isovolumetric process, which as you may recall tells us that the delta e internal to change in internal energy is simply the heat as in that the work is zero so that we know the work from b to c and the work from d to a are zero as in the work from a to c is really just the work from a to b to c and the work from c d to a is really just the work from c to d okay that's all very well and good but what is our work from c to d so let's try and calculate it using the above expression, work from c to d equals negative p, and this is p c, d because we're at that level.

03:28 And then we're going from volume, the volume at, the final volume we're going to is the volume at points a and d, and that's minus the volume from points b and c, which is our initial volume, even though it's larger.

03:42 So this is going to be a negative quantity.

03:45 But we don't have a value for this quantity.

03:47 So let's try and evaluate it from a similar quantity that we do have the value for, specifically, the work from a to b, which we know is given by negative 5 p sub cd times.

04:01 Again, this is the volume from b to c minus the volume from a, d.

04:10 And we know that this equals negative 500 joules.

04:14 So we can go along and simplify.

04:17 This that to get to that negative p sub cd times just the this vbc minus vbc minus v a d is equal to sorry we're going to cancel out this negative too because we're just going to divide both sides by negative five and we would get that p sub cd times vbc minus v a d is just equal to a hundred joules...

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