An ideal gas initially at Pi, Vi, and Ti is taken through a cycle as shown in Figure P 20.34 . (

An ideal gas initially at Pi, Vi, and Ti is taken through a...

Key Concepts

Internal Energy of an Ideal Gas

For an ideal gas, the internal energy is a function only of temperature and is independent of pressure and volume. This concept simplifies the analysis of thermodynamic cycles, as any change in the internal energy of an ideal gas can be directly related to a change in temperature.

Ideal Gas Law

The ideal gas law is a fundamental equation, PV = nRT, that relates the pressure, volume, and temperature of an ideal gas. It is crucial for understanding how gases respond to changes in any one of these variables and forms the basis for analyzing gas processes in various thermodynamic cycles.

Thermodynamic Cycle

A thermodynamic cycle is a sequence of processes that returns a system to its original state, meaning that the internal energy change over one complete cycle is zero. The net work done by or on the system over the cycle is related to the area enclosed by the process path on a pressure-volume (PV) diagram.

Pressure-Volume Work

Pressure-volume work refers to the work done by a gas when it expands or contracts under a given pressure, calculated as the integral of pressure with respect to volume. In cyclic processes, the net work done is determined by the closed-loop integral over the cycle, which is the area enclosed on a PV diagram.

First Law of Thermodynamics

The first law of thermodynamics is a statement of energy conservation, asserting that the energy change in a system equals the heat added to the system minus the work done by the system. This principle is essential for analyzing energy exchanges in any thermodynamic process and for determining net energy added or removed during a cycle.

Transcript

00:01 Consider an ideal gas that goes through the process as depicted by the figure on the left, starting at point a with pressure, p -s -i, volume, v -s -i, and temperature t -s -i.

00:14 And it goes from a to b, b to c, c to d, and d to a.

00:18 And the first thing we're wondering about this ideal gas is what is the network done on it per cycle? and a handy trick is that we know for cyclic processes, the area enclosed, by the path of the process on pv pressure times volume diagram is equal to the network per cycle.

00:38 So this is actually the fastest way to do this is just literally the area of this square.

00:44 So the height free pi minus 2pi or free pi minus pi of just to p.

00:53 And times the width of 3 v sub i minus v sub i again to v.

00:58 And due to the direction of the path this is going to be negative because you're going to be and then b to c at free p i and then c the d and then d to a at p i so this is going to be just negative for p i v i is equal to the word per cycle and that's the fastest simplest way of doing that but i'm going to show it just in case you don't believe me um so we can just use our classic equation 20 .9 to give us that the work is equal to negative times the integral of the initial volume to the final volume of the pressure over dv and we're gonna break this into parts so first is the integral from a to b which is vi to the or vi to the volume stays constant and the pressure doesn't do anything.

02:10 There's no work done here.

02:12 We know that for any isoeuvometric process where the volume is constant, the change in internal energy is just a change in heat.

02:19 So we know there's no work done on this little pdv leg for a to b.

02:25 So we can go ahead and say that's zero.

02:29 And then we're looking at the leg from b to c and this i'm just distributing this negative to each piece.

02:38 So this time we're going from volume vi to three vi and the pressure is a constant three piece of i.

02:45 We're going to integrate that over dv and then we're going to add our leg from c to d which just like our leg from a to b, just like our leg from a to b is an isometric process meaning that the only change in internal energy is the heat change.

03:05 There's no work done here.

03:08 So again, going from free v .i to three v.

03:11 Of this pdv is, we can go ahead and say zero again, because even if it was something, it would just be canceling itself just from the integral, even outside of just thinking of the physics.

03:26 And then finally, from d to a, of from three vi, v .i to i to v .i.

03:33 And now, that's the constant pressure of just p.

03:35 I v v.

03:37 So we can go ahead and evaluate our two remaining integrals.

03:42 To get that v evaluates to negative three piece of i of just this got constant carrying through and then it's just v from v subi to free v sub i.

04:01 So free v.

04:03 I minus v sub i minus and then from this integral just the constant p sub i and this time we're going from free v sub i to v sub i so just v sub i minus free v sub i and that just simplifies to negative four p sub i v so i so we didn't need to do the integral but i just wanted to show you that that trick of the area enclosed by the path of the process does work, but this confirms that the work done is negative 4 p sub i v sub i.

04:37 So the next question might be what is to change in heat in our ideal gas and recall that our sick, this same concept of the cyclic process being the area enclosed by the path of process on the diagram is mathematically the same as saying that the delta e internal is equal to zero in that namely q equals negative w as in if we go back to this diagram the work done in b to some c and a sub d is equal to the negative magnitude of the heat acquired in a sub b and c d because what's going on here like a to b the volume is constant while the pressure expands from the ideal gas law pv equals nrt we know the temperature is going up there.

05:34 So the q is being added here...

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