Consider a process carried out on 1.00 mol of a monatomic ideal gas by the following two different pathways. The first pathway is A (3.00 atm, 20.0 L) to C (1.00 atm, 20.0 L) to D (1.00 atm, 50.0 L) ; and the second pathway is A (3.00 atm, 20.0 L) to B (3.00 atm, 50.0 L) to D (1.00 atm, 50.0 L) . In each case, the gas is taken from state A to state D.
-Calculate w_AC.

Question

Quizplus · Accepted Answer

The work done on a gas during an isothermal process is given by $w = -nRT \ln\left(\frac{V_f}{V_i}\right)$ for an ideal gas, where $n$ is the number of moles, $R$ is the gas constant, $T$ is the temperature, $V_f$ is the final volume, and $V_i$ is the initial volume. However, in this question, we are not directly calculating work using this formula but rather understanding the concept of work in thermodynamics. The work done on a gas is path-dependent, but when considering a cyclic process or comparing paths between the same initial and final states, the internal energy change ($\Delta U$) of the system is zero because internal energy is a state function and depends only on the initial and final states, not on the path taken. Since the question involves calculating work done ($w$) and given that both pathways return to the same initial and final states, the total change in internal energy ($\Delta U$) for the entire process (from A to D through any path) is zero because internal energy is a state function. The question seems to be asking for a calculation or a concept that is not fully explained in the prompt (as indicated by the placeholder "w_AC"), but based on the principles of thermodynamics, the key takeaway is that the change in internal energy for a cyclic process or any process that returns to the initial state is zero, regardless of the path taken.

Consider a Process Carried Out on 1