Animated Solution for Chemistry - Chemical Thermodynamics: A reversible cyclic process for an ideal gas is shown below. Here, P , V and T are pressure , volume and temperature , respectively. The thermodynamic parameters q, w, H and U are heat, work, enthalpy and internal energy, respectively.

Welcome, future engineers and doctors! Today, we are going to dissect a fascinating problem from JEE Advanced 2018. This question is a beautiful amalgamation of graphical analysis and the First Law of Thermodynamics. It tests not just your memory of formulas, but your deep conceptual understanding of how state variables interact during different thermodynamic processes.

The V-T Graph

Imagine you are looking at the heartbeat of an ideal gas. The graph provided is a Volume-Temperature (V-T) plot. Before we even look at the options, our first mission is to decode the paths. A common mistake students make is rushing to the equations without fully understanding the physical journey the gas is taking. Let's break it down step-by-step.

We have three distinct points: $A(P_1,V_1,T_1)$, $B(P_2,V_2,T_1)$, and $C(P_2,V_1,T_2)$.

Process A $\to$ B: Look at the line connecting A and B. It's perfectly vertical. On a V-T graph, a vertical line means the temperature is constant. Thus, this is an isothermal process. The gas is being compressed from a larger volume $V_1$ to a smaller volume $V_2$ while maintaining a constant temperature $T_1$.

Process B $\to$ C: This path is a straight line that, if extended, would pass through the origin. What does this tell us? It screams Charles's Law! Volume is directly proportional to temperature ($V\propto T$), which only happens when the pressure is constant. Therefore, this is an isobaric process at pressure $P_2$. The gas expands from $V_2$ back to $V_1$ as it heats up from $T_1$ to $T_2$.

Process C $\to$ A: Finally, the gas returns to its initial state via a horizontal line. A horizontal line on this graph means the volume is locked at $V_1$. This is an isochoric process. The gas cools down from $T_2$ to $T_1$ at a constant volume.

Now that we have our roadmap, let's test the options.

Option A suggests that the work done in process A $\to$ B is $P_2(V_2-V_1)$. But wait! We just established that A $\to$ B is an isothermal process. The work done in a reversible isothermal process involves a natural logarithm: $w_{AB}=-nR{T}_1\ln \left(\frac{V_2}{V_1}\right)$. It is definitely not a simple $P\Delta V$ calculation. So, Option A is out of the game.

Let's move to Option B. It claims $w_{BC}=P_2(V_2-V_1)$. Let's calculate it ourselves. Process B $\to$ C is isobaric. In chemistry, the IUPAC sign convention for work is $w=-P_{ext}\Delta V$. Substituting our values: If we distribute the negative sign inside the bracket, we get: Brilliant! The first part of Option B is correct.

What about the second part? It says $q_{BC}=\Delta H_{AC}$. For the isobaric process B $\to$ C, the heat exchanged is equal to the change in enthalpy: Now, what is the enthalpy change for the hypothetical process A $\to$ C? Since enthalpy is a state function, it only depends on the initial and final temperatures. They match perfectly! Option B is a correct choice.

Option C presents a very interesting inequality: $\Delta H_{CA}<\Delta U_{CA}$. Let's analyze the cooling process from C to A. The temperature drops from $T_2$ to $T_1$. The change in enthalpy is: The change in internal energy is:

Since $T_1$ is less than $T_2$, the term $(T_1-T_2)$ is negative. We also know that the molar heat capacity at constant pressure ($C_P$) is always greater than the molar heat capacity at constant volume ($C_V$) for an ideal gas. When you multiply a larger positive number ($C_P$) by a negative number, you get a more negative result than multiplying a smaller positive number ($C_V$) by that same negative number. In the realm of mathematics, a more negative number is strictly less than a less negative number. Therefore: This is a classic trap! Many students see $C_P>C_V$ and instinctively assume $\Delta H>\Delta U$, forgetting to account for the negative sign during cooling.

The second part of Option C states $q_{AC}=\Delta U_{BC}$. Process A $\to$ C is isochoric, so the heat exchanged is: The internal energy change for process B $\to$ C is: Once again, a perfect match! Option C is also correct.

Finally, Option D claims $\Delta H_{CA}>\Delta U_{CA}$. Since we just rigorously proved the exact opposite, Option D is incorrect.

By carefully decoding the V-T graph and applying the First Law of Thermodynamics with strict adherence to sign conventions, we have successfully navigated this problem. The correct options are indeed (B) and (C). Remember, in thermodynamics, the graph is your map, and the state functions are your compass. Trust them, and you won't get lost!