Animated Solution for Chemistry - Chemical Thermodynamics: Which one of the following equations does not correctly represent the first law of thermodynamics for the given processes involving an ideal gas? (Assume non-expansion work is zero)

Decoding Process Equations

Thermodynamics is the beautiful study of energy, heat, and work. At the very heart of this subject lies the First Law of Thermodynamics, which is essentially the law of conservation of energy applied to a thermodynamic system.

Mathematically, the first law is expressed as:

Here, $\Delta U$ represents the change in the internal energy of the system, $q$ is the heat exchanged between the system and its surroundings, and $W$ is the work done on or by the system.

Before we dive into the specific processes, it is crucial to understand the IUPAC sign convention. Heat added to the system is positive ($+q$), and heat lost by the system is negative ($-q$). Similarly, work done on the system (compression) is positive ($+W$), while work done by the system (expansion) is negative ($-W$).

Let's evaluate how the First Law adapts to different thermodynamic constraints.

1. Cyclic Process In a cyclic process, a system undergoes a series of changes but ultimately returns to its exact initial state. Because internal energy ($U$) is a state function, its overall change over the cycle is zero.

Substituting this into the First Law gives $0=q+W$, which rearranges to $q=-W$. This perfectly matches option (a).

2. Adiabatic Process An adiabatic process is defined by perfect thermal insulation. The system can neither absorb nor release heat, meaning $q=0$.

Substituting $q=0$ into the First Law yields:

However, option (b) claims that $\Delta U=-W$. This is a direct violation of the First Law under adiabatic conditions.

3. Isochoric Process In an isochoric process, the volume of the system remains strictly constant ($\Delta V=0$). Since pressure-volume work is given by $W=-p_{ext}\Delta V$, the work done is zero ($W=0$).

Plugging this into our master equation leaves us with $\Delta U=q$. This confirms that option (c) is correct.

4. Isothermal Process For an isothermal process involving an ideal gas, the temperature remains constant ($\Delta T=0$). The internal energy of an ideal gas is solely a function of its temperature. Therefore, no change in temperature means no change in internal energy ($\Delta U=0$).

Just like in the cyclic process, this leads to $0=q+W$, or $q=-W$. Option (d) is also correct.

By systematically applying the constraints of each process to the First Law of Thermodynamics, we easily spotted the imposter. Option (b) incorrectly states the relationship for an adiabatic process. Always trust the master equation and the sign conventions, and these problems will become second nature!