# PV diagram-calculating work

• M_CG
In summary, the question involves finding the change in internal energy of a gas that expands from I to F along a diagonal path, with a given energy added by heat. The attempt at a solution involves using the equations for (delta)U=Q+W and W=P(delta)V, but the answer was incorrect due to incorrect units. After correcting for units and average pressure, the final answer is -120.8125 J. The question of atomicity of the gas is raised, but remains unsolved.

## Homework Statement

http://i41.tinypic.com/r7rq5w.jpg
The question is stated in the picture but I'll restate it:
A gas expands from I to F in the figure. The energy added to the gas by heat is 449 J when the gas goes from I to F along the diagonal path. What is the change in internal energy of the gas? Answer i units of J.

## Homework Equations

(delta)U=Q+W (In this case W is negative because the gas is doing work on the system)
W=P(delta)V

## The Attempt at a Solution

Since Q is given, I need to find W done by the gas.
Pressure is not constant, but its change is constant so I can use Paverage
Paverage=2*1.013*105Pa
Therefore W=(2*1.013*105)*2.5L=506,500 Joules
Inputting that into the equation for (delta)U:
449 J-506,500 J= -506,051 J

However this answer was incorrect. I also tried calculating work by finding the area under the graph IF, but this was wrong as well. I'm not really sure what I'm doing wrong. Could someone point me in the right direction?

To get work in Joules (=Nm), you want volume in m^3 not litres, and P in Pa (=N/m^2).

An interesting follow up question would be "how many atoms in each molecule of the gas?"

Last edited:
Thank you! I had a feeling there was something funny going on with my units. I also figured out that I was calculating average pressure incorrectly.
W=(2.25*1.013*105 pa)(.0025m3)=569.8125 J
$$\Delta$$U=449 J-569.8125 J=-120.8125 J

Any idea about the atomicity of the gas?

Ahh...that question might be a little over my head. I don't know how to go about figuring that one out.

## 1. What is a PV diagram?

A PV diagram is a graphical representation of the relationship between pressure and volume in a thermodynamic system. It is used to visualize and analyze the changes in a system's internal energy as it undergoes a process.

## 2. How do you calculate work from a PV diagram?

To calculate work from a PV diagram, you need to find the area under the curve on the graph. This can be done by dividing the graph into smaller shapes, such as rectangles or triangles, and calculating the area of each shape. Then, you can add up the areas of all the shapes to get the total work done.

## 3. What is the significance of calculating work from a PV diagram?

Calculating work from a PV diagram can help us understand the energy transfer that occurs during a process. It is also useful in determining the efficiency of a system and evaluating the performance of different thermodynamic processes.

## 4. What are the units of work in a PV diagram?

The units of work in a PV diagram are joules (J) or newton-meters (N·m). These units represent the amount of energy transferred during a process.

## 5. Can a PV diagram be used to calculate work for all thermodynamic processes?

No, a PV diagram can only be used to calculate work for processes that involve a change in volume at a constant pressure. Other types of processes, such as isothermal or adiabatic processes, require different methods to calculate work.