Imagine you are a tiny observer inside a cylinder of an ideal gas, watching the molecules bounce around as they undergo a fascinating thermodynamic journey. This problem asks us to track the evolution of five critical state variables—pressure (p), volume (V), temperature (T), enthalpy (H), and entropy (S)—across two distinct reversible processes.
Our mission is to play detective and determine which of the four provided graphs accurately depict this journey. Let's break it down step by step.
Leg 1
The Isothermal Expansion (State I to II)
The first leg of our journey is a reversible isothermal expansion. The word 'isothermal' is our biggest clue here: it means the temperature (T) remains absolutely constant throughout the process.
Because the gas is expanding, its volume (V) is increasing. According to Boyle's Law, for a fixed amount of ideal gas at a constant temperature, pressure and volume are inversely proportional (pV=constant). Therefore, as the volume goes up, the pressure (p) must go down.
Now, let's talk about the energy state. For an ideal gas, the internal energy and the enthalpy (H) are purely functions of temperature. Since the temperature isn't changing, the enthalpy remains perfectly constant (H=constant).
What about entropy (S)? Entropy is a measure of the system's randomness or the number of available microstates. As the gas expands into a larger volume, the molecules have more space to move around. This increased freedom means the randomness increases, and thus, the entropy increases (S↑).
Leg 2
The Adiabatic Expansion (State II to III)
Next, the gas undergoes a reversible adiabatic expansion. 'Adiabatic' means there is no heat exchange with the surroundings (q=0). Because the process is also reversible, the change in entropy is zero (ΔS=∫Tdqrev=0). This makes the process isentropic, meaning the entropy remains strictly constant (S=constant).
The gas is still expanding, so the volume (V) continues to increase. However, because no heat is entering the system to fuel this expansion, the gas must do work at the expense of its own internal energy. As the internal energy depletes, the temperature (T) drops.
Since we established earlier that the enthalpy of an ideal gas depends directly on its temperature, a drop in temperature guarantees a drop in enthalpy (H↓).
Finally, what happens to the pressure? The pressure drops for two compounding reasons: the volume is increasing, and the temperature is decreasing. This dual effect causes the pressure to drop even more sharply than it did during the isothermal phase.
Decoding the Graphs
Armed with our thermodynamic logic, let's evaluate the four options.
Graph (A): The p−V Plot
From state I to II, the pressure decreases as volume increases along a standard isothermal curve. From state II to III, the pressure drops further as volume increases, but this time along a steeper adiabatic curve. This perfectly matches our analysis. Graph (A) is correct.
Graph (B): The p−T Plot
From state I to II, the temperature is constant while the pressure drops, which visually translates to a vertical line heading downwards. From state II to III, both the temperature and the pressure decrease, forming a curve that moves downwards and to the left. The graph depicts exactly this behavior. Graph (B) is correct.
Graph (C): The H−S Plot
From state I to II, enthalpy is constant while entropy increases, represented by a horizontal line moving to the right. So far, so good. However, from state II to III, entropy is constant while enthalpy should decrease (a vertical line moving downwards). The provided graph shows the vertical line moving upwards, implying an increase in enthalpy. This is a trap! Graph (C) is incorrect.
Graph (D): The T−S Plot
From state I to II, temperature is constant and entropy increases, shown as a horizontal line to the right. From state II to III, entropy is constant and temperature decreases, shown as a vertical line moving downwards. This aligns flawlessly with our theoretical predictions. Graph (D) is correct.
By carefully tracking the state variables through each thermodynamic process, we can confidently conclude that the correct plots are A, B, and D.
