Calculating the work done on a decomposed solid

[laxgal12]

Homework Statement

A sample consisting of 1.0 mol CaCO3 (s) was heated to 800 degrees Celsius, when it decomposed. The heating was carried out in a container fitted with a piston that was initially resting on the solid. (1) Calculate the work done during complete decomposition at 1.0atm. (2) What work would be done if insead of having a piston the container was open to the atmosphere?

Homework Equations

For (2), w_{constant external pressure}= -p_external$$\Delta$$T

The Attempt at a Solution

I'm just not clear about what's happening in this situation...when the CaCO3 is decomposing, I assume the volume is changing...but is the volume larger than when it was in its solid form, or smaller (due to the presence of the piston). Is the pressure on the piston changing? To what does the 1.0 atm refer to- is that a constant pressure on the piston, or only refer to the pressure at which the decomposition takes place, and then th pressure changes?
Once I understand what's actually happening with regard to this decomposition, I can select the correct equations, so I don't really need anyone to 'solve' this problem for me- I just am having difficulty understanding what's changing, what's staying the same, in what direction the change is occurring, etc.

Thank you very much for your time and effort =)

 

[Borek]

Start with the reaction of carbonate decomposition, that should give you a hint.

 

[Blueshift5]

it is important to have a clear understanding of the system and variables involved in order to accurately calculate the work done. In this case, the system is a sample of CaCO3 (s) undergoing decomposition at a constant pressure of 1.0 atm. The container is fitted with a piston, which means that the volume of the system is constant. The temperature is the only variable that is changing, as the sample is heated to 800 degrees Celsius.

For (1), we can use the equation w = -PΔV to calculate the work done. Since the volume is constant, the work done is equal to the change in pressure multiplied by the change in volume. In this case, the volume remains constant, so the work done is 0.

For (2), we can use the equation w = -PΔT to calculate the work done. Since the container is open to the atmosphere, the pressure is not constant, but the temperature change is the same as in (1). Therefore, the work done would be equal to the change in pressure (from 1.0 atm to atmospheric pressure) multiplied by the change in temperature (from 800 degrees Celsius to room temperature).

In summary, the work done in both cases is dependent on the change in pressure and temperature. However, the presence of the piston in (1) results in a constant volume, while the lack of a piston in (2) allows for a change in volume. It is important to consider all variables and their effects on the system in order to accurately calculate the work done.