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

(a) No work is performed during the expansion of the gas. (b) This is because there is no external force opposing the gas expansion, and the gas simply flows from the filled flask to the evacuated one without exerting force on any object or boundary. (c) Yes, we can determine the value of \(\Delta E\) for the process, and it is 0 due to no heat transfer and no work done on/by the system.

Step-by-step solution

Understand the concept of work in thermodynamics

In thermodynamics, work is done on or by a system undergoing a transformation. When a gas expands, it can perform work on its surroundings. However, not all expansions result in work done. The key factor determining whether work is done is whether there is an opposing force against the expansion. Commonly, work is done in systems with external pressure (e.g., a piston in a cylinder).

Analyze the given system

In the given system, there is no external pressure or force opposing the gas expansion, as the gas expands into the evacuated flask. As a result, it automatically fills the available empty space without exerting force on any object or screen.

Answering (a) and (b)

(a) No work is performed during the expansion of the gas. (b) It's because there is no external force opposing the gas expansion, and the gas simply flows from the filled flask to the evacuated one without exerting force on any object or boundary.

Apply the first law of thermodynamics

The first law of thermodynamics states that \(\Delta E = Q - W\) where \(\Delta E\) is the change in internal energy of the system, \(Q\) is the heat transfer, and \(W\) is the work done on/by the system. Since the system is perfectly insulated, no heat transfer occurs (\(Q = 0\)), and no external work is done (\(W = 0\)).

Solving for \(\Delta E\) and answer (c)

Given \(Q = 0\) and \(W = 0\), using the first law of thermodynamics, we have: \(\Delta E = 0 - 0 = 0\). (c) Yes, we can determine the value of ΔE for the process, and it is 0 due to no heat transfer and no work done on/by the system.