Work Done in Ideal Gas System

In summary, the problem involves determining the net work done by an ideal gas during a three-part process, with given pressures and volumes. The work done from C to A is zero, and from B to C is -150kJ. The work done from A to B is 150kJ, giving a total work of zero. However, the correct answer is actually 300kJ, as the work done from A to B includes the area under the entire graph, not just the purple triangle.
  • #1
eriadoc
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Homework Statement



An ideal gas follows the three-part process shown in the figure.

Walker4e.ch18.Pr025.jpg


At the completion of one full cycle, find the net work done by the system.

Homework Equations



W=P*deltaV; Total Work = Wab+Wbc+Wca
A=1/2bh

The Attempt at a Solution



Work done from C to A is zero, because the volume is not changing. Work done from B to C is (50kPa)(-3) = -150kJ.

Work done from A to B is where I'm lost. I know it's supposed to be the area, so I did .5(100kPa)(3) = 150kJ.

Therefore, total work would be zero. The answer is 150kJ, but I can't figure out why. I've read the answer on Cramster and it honestly confused me even more. I feel like I'm close, but I just need an explanation more so than the math. TIA.
 
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  • #2
It's not the area of the triangle you're looking for, it's the area under the graph to the V-axis you want. See it as a basic integral W = a∫bpdV. To put it more simply, the work done by one process is that integral. Since you're looking for the total work done by the system, all you really need to do is calculate the area of the triangle as it already has subtracted Wbc.
 
Last edited:
  • #3
da nang is right. The work done by the gas from A to B is the whole area under AB not just the purple triangle - ie. the purple triangle PLUS the area under BC = 300 KJ.

AM
 

1. What is work done in an ideal gas system?

In an ideal gas system, work is defined as the energy transferred to or from the system through a force acting on the system. This can be in the form of pressure-volume work, where the gas expands or is compressed, or non-expansion work, such as stirring or agitation.

2. How is work calculated in an ideal gas system?

The work done in an ideal gas system can be calculated using the equation W = -PΔV, where W is the work done, P is the pressure, and ΔV is the change in volume of the gas. This equation applies for both expansion and compression work.

3. What is the relationship between temperature and work done in an ideal gas system?

According to the ideal gas law, PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the gas constant, and T is the temperature. As temperature increases, the volume of the gas also increases, resulting in more work done in an expansion process. Conversely, as temperature decreases, the volume decreases, resulting in less work done in a compression process.

4. What is the difference between work done in an ideal gas system and a real gas system?

In an ideal gas system, the gas particles are assumed to have no volume and do not interact with each other, resulting in a simpler calculation for work. In a real gas system, the gas particles do have volume and may attract or repel each other, making the calculation of work more complex. Real gas systems also take into account the non-ideal behavior of gases at high pressures and low temperatures.

5. Can work be done on an ideal gas system without changing its temperature?

Yes, work can be done on an ideal gas system without changing its temperature. This is known as isothermal work, where the temperature of the gas remains constant throughout the process. In this case, the work is calculated using the equation W = -nRTln(V2/V1), where V2 and V1 are the initial and final volumes of the gas, respectively.

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