Question

Consider a system consisting of the following apparatus, in which gas is confined in one flask and there is a vacuum in the other flask. The flasks are separated by a valve. Assume that the flasks are perfectly insulated and will not allow the flow of heat into or out of the flasks to the surroundings. When the valve is opened, gas flows from the filled flask to the evacuated one. (a) Is work performed during the expansion of the gas? (b) Why or why not? (c) Can you determine the value of \(\Delta E\) for the process?

Short answer

In summary, for this adiabatic process involving gas expansion into a vacuum: (a) No work is performed during the expansion since the gas doesn't have to push against any external pressure. (b) This explains why no work is performed during the process. (c) Consequently, the change in internal energy \(\Delta E\) for the process is 0.

Step-by-step solution

Understanding the type of process

The initial state has gas confined in one flask and a vacuum in the other, separated by a valve. It is given that the flasks are perfectly insulated, meaning this is an adiabatic process - there is no exchange of heat between the system (gas) and its surroundings.

Determining Work Performed

In an adiabatic process, no heat exchange occurs between the system and its surroundings. When the valve is opened, the gas expands from the filled flask to the evacuated one. However, during this expansion process, the gas is not pushing against any external pressure, as it just fills the vacuum in the other flask. Therefore, no work is performed during the expansion of the gas.

Explaining Why No Work Is Performed

Work is performed when a force acts over a certain displacement. In our case, when the gas expands, it does not have to overcome any force or push against any external pressure since the other flask is initially a vacuum. As a result, there is no work done during the expansion of the gas.

Determining the Change in Internal Energy (ΔE)

For an adiabatic process, the first law of thermodynamics states that the change in internal energy, \(\Delta E\), is equal to the work done on the system (\(W\)). In our case, we concluded in step 3 that no work is performed during the expansion. Therefore, the change in internal energy is: \[ \Delta E = W \] In this situation, we have \(W = 0\), which means the change in internal energy is also zero: \[ \Delta E = 0 \] So, the change in internal energy for this process is zero. To conclude, (a) No work is performed during the expansion of the gas. (b) This is because the gas is expanding into a vacuum and doesn't have to push against any external pressure. (c) The value of \(\Delta E\) for the process is 0.