One mole of an ideal gas is taken through the cyclio process shown in Fig. 12.24 . (a) Compute t

One mole of an ideal gas is taken through the cyclio process...

Key Concepts

Cyclic Processes

A cyclic process is one in which the system returns to its initial state after completing a series of processes. Because state functions, like internal energy, depend only on the state of the system, their net change over a complete cycle is zero. However, work and heat, being path-dependent, can be nonzero over the cycle.

Path Functions vs. State Functions

State functions, such as pressure, volume, temperature, and internal energy, depend solely on the current state of the system, irrespective of how it arrived there. In contrast, path functions like work and heat rely on the specific process taken between states. This distinction is vital when analyzing energy exchanges in cyclic and other thermodynamic processes.

Work in Thermodynamic Processes

Work in a thermodynamic process is calculated by integrating the pressure with respect to volume (W = ?PdV). The value and sign of the work depend on the specific nature of the process (e.g., isothermal, adiabatic, isobaric, or isochoric) and the path taken during the process.

Ideal Gas Law

The Ideal Gas Law, expressed as PV = nRT, relates the pressure, volume, and temperature of an ideal gas. It provides the basis for understanding how these variables interact during different thermodynamic processes and is crucial for resolving problems involving ideal gas cycles.

Internal Energy of Ideal Gases

For an ideal gas, internal energy depends solely on temperature, not on volume or pressure. This means that any change in internal energy can be computed directly from the temperature change using the relation ?U = nCv?T, where Cv is the molar heat capacity at constant volume.

First Law of Thermodynamics

This fundamental principle states that the change in a system's internal energy is equal to the difference between the heat added to the system and the work done by the system (?U = Q - W). It is essential for analyzing energy transfers during any thermodynamic process, including cyclic ones.

Transcript

00:01 So we have some pressures and volumes that are shown for a particular heat engine cycle using one mole of an ideal gas.

00:12 And we need to find work done by this gas for the various segments in the rectangular cycle.

00:25 So work done is equal pressure times the change in volume.

00:35 We have the pressures.

00:37 Just for reference, one atmosphere, we're going to be converting that into the si units of pascal's.

00:49 A known conversion factor.

00:52 But we don't know the volumes, so we will have to use the ideal gas law at some of the points with known temperatures in order to figure out the volume.

01:03 Okay, so volume 2, what i'll need is i'll need point a.

01:14 So pressure at a times volume 2 is equal to one mole times the gas constant.

01:26 Oh, i should write down the ideal gas law.

01:33 Pv equals n -r -t, where n is the number of moles and r is the gas constant.

01:43 8 .31 joules per mole kelvin.

01:53 Okay, so we have n as 1, of course, 8 .31.

01:58 And our temperature at point a, they show some isotherms, so we can read off the temperature as 400 kelvin, and the units should come out to be joules.

02:12 So we have to divide by the pressure, and i'll have to convert that into my si units so that my volume comes out to be in cubic meters.

02:39 And we have one atmosphere and then put in the conversion into pascal's.

02:59 So one of our unknown volumes pops out and that's a good start.

03:08 Okay, our second volume, we can then see that since the pressure stayed the same, so we're going to kind of use the ideal gas law, but in a way that's a little bit maybe cheating.

03:28 Well, not cheating, but simplifying.

03:31 So pressure at point b equals the pressure at a, but temperature at b is one -half, temperature at a.

03:50 So what that tells us is the volume one is one -half, volume two.

03:57 We could put it into the ideal gas law, but why go to all that work? only if you felt like you were not doing things right.

04:11 Okay, so now we can figure out the work done in each of the segments because we have the volumes.

04:20 So the easy ones are the straight up and down lines.

04:24 So b to c and d to a, no work.

04:36 No change in volume...