Solving Problems with Non-Standard PV Paths

Hello! Welcome to your twelfth lesson.

In our previous session, we developed a robust technique for analyzing processes that follow linear paths on P-V, V-T, and P-T diagrams. You mastered calculating thermodynamic quantities for these paths and, critically, applied calculus to find the maximum temperature reached during a process—a key skill from your math background.

Today, we take the final step in analyzing process paths. We will move from straight lines to curves. The learning outcome for this lesson is to solve problems involving non-standard curved process paths on P-V diagrams (e.g., $P = α V^{2}$ ).

You'll find that the calculus-based methods from the last lesson are not just relevant; they are the central tools we'll use today. We'll be applying integration and differentiation to handle any process defined by a mathematical function, a common feature in JEE Advanced problems.

1. The General Approach for Any Process Path

Let's start with the first principles. For any thermodynamic process, the work done by the gas is defined by the integral:
$W = \int_{V_{i}}^{V_{f}} P d V$

In previous lessons, we saw how this integral simplifies for standard processes:

Isobaric ( $P = const$ ): $W = P (V_{f} - V_{i})$
Isothermal ( $P = n R T / V$ ): $W = n R T ln (V_{f} / V_{i})$
Linear Path ( $P = mV + c$ ): $W = Area of trapezoid$

When a process follows a non-standard curve, like $P = α V^{2}$ or $P = P_{0} e^{- V / V_{0}}$ , there are no simple geometric shortcuts. We must return to the definition and solve the integral directly. This is where your calculus skills become a significant advantage.

Once work ( $W$ ) is found, the rest of the analysis follows the familiar pattern:

Find endpoint states: Determine $(P_{i}, V_{i}, T_{i})$ and $(P_{f}, V_{f}, T_{f})$ using the process equation and the ideal gas law.
Calculate $Δ U$ : Use $Δ U = n C_{v} Δ T$ . Since internal energy is a state function, this only depends on the initial and final temperatures.
Calculate $Q$ : Apply the First Law of Thermodynamics, $Q = Δ U + W$ .

Let's apply this systematic approach to a past JEE Main problem.

2. Worked Example: Analyzing a Non-Standard Path

Consider the following problem, which involves a process defined by a specific non-linear equation. We will break it down step-by-step to calculate all the relevant thermodynamic quantities.

Solved papers for JEE Main & Advanced Physics Thermodynamical ...

This problem is from a previous JEE Main paper. It asks for the change in temperature for a specific non-standard process. Let's study the problem statement.

Read Question 72. An ideal gas undergoes a process where pressure and volume are related by the equation: P = P_0 \left[ 1 - \frac{1}{2} \left( \frac{V_0}{V} \right)^2 \right] The question asks to find the change in temperature as the gas volume changes from V_0 to 2V_0.

This problem is a perfect test of our general strategy.

Step 1: Calculate the Change in Temperature ( $Δ T$ )

The question asks for $Δ T = T_{f} - T_{i}$ . We can find the initial and final temperatures using the ideal gas law, $T = P V / (n R)$ .

Initial State (i): $V_{i} = V_{0}$
Pressure $P_{i} = P_{0} [1 - \frac{1}{2} (\frac{V _{0}}{V _{0}})^{2}] = P_{0} [1 - \frac{1}{2}] = \frac{1}{2} P_{0}$ .
Temperature $T_{i} = \frac{P _{i} V _{i}}{n R} = \frac{( \frac{1}{2} P _{0} ) V _{0}}{n R} = \frac{P _{0} V _{0}}{2 n R}$ .
Final State (f): $V_{f} = 2 V_{0}$
Pressure $P_{f} = P_{0} [1 - \frac{1}{2} (\frac{V _{0}}{2 V _{0}})^{2}] = P_{0} [1 - \frac{1}{8}] = \frac{7}{8} P_{0}$ .
Temperature $T_{f} = \frac{P _{f} V _{f}}{n R} = \frac{( \frac{7}{8} P _{0} ) ( 2 V _{0} )}{n R} = \frac{7 P _{0} V _{0}}{4 n R}$ .
Change in Temperature:
$Δ T = T_{f} - T_{i} = \frac{P _{0} V _{0}}{n R} (\frac{7}{4} - \frac{1}{2}) = \frac{P _{0} V _{0}}{n R} (\frac{7 - 2}{4}) = \frac{5}{4} \frac{P _{0} V _{0}}{R}$
(Assuming $n = 1$ mole as per the convention in many such problems). This matches option (C) in the resource.

Step 2 (Extended Analysis): Calculate Work Done ( $W$ )

Let's go beyond the question and calculate the work done, which requires direct integration.
$W = \int_{V_{0}}^{2 V_{0}} P (V) d V = \int_{V_{0}}^{2 V_{0}} P_{0} [1 - \frac{1}{2} \frac{V _{0}^{2}}{V ^{2}}] d V$
$W = P_{0} [V - \frac{1}{2} V_{0}^{2} (- \frac{1}{V})]_{V_{0}}^{2 V_{0}} = P_{0} [V + \frac{V _{0}^{2}}{2 V}]_{V_{0}}^{2 V_{0}}$
$W = P_{0} [(2 V_{0} + \frac{V _{0}^{2}}{4 V _{0}}) - (V_{0} + \frac{V _{0}^{2}}{2 V _{0}})]$
$W = P_{0} [(2 V_{0} + \frac{V _{0}}{4}) - (V_{0} + \frac{V _{0}}{2})] = P_{0} V_{0} (\frac{9}{4} - \frac{3}{2}) = \frac{3}{4} P_{0} V_{0}$

With $W$ and $Δ T$ , you could now easily find $Δ U$ and the total heat $Q$ .

3. Optimization on a Curved Path: Finding Maximum Temperature

In the last lesson, we found the maximum temperature for a linear process. The same optimization technique applies to curved paths.

Let's analyze a common problem type: A process is described by $P = P_{0} - α V^{2}$ , where $P_{0}$ and $α$ are positive constants. Find the maximum temperature the gas can attain.

Express T as a function of one variable (V):
From the ideal gas law, $T = \frac{P V}{n R}$ . Substituting the process equation:
$T (V) = \frac{( P _{0} - α V ^{2} ) V}{n R} = \frac{1}{n R} (P_{0} V - α V^{3})$
Find the extremum using calculus:
To find the volume $V_{ma x T}$ at which temperature is maximum, we differentiate $T (V)$ and set the derivative to zero:
$\frac{d T}{d V} = \frac{1}{n R} (P_{0} - 3 α V^{2}) = 0$
Solving for $V$ , we get:
$V_{ma x T} = \frac{P _{0}}{3 α}$
(We take the positive root as volume is positive).
Calculate the maximum temperature:
Substitute $V_{ma x T}$ back into the expression for $T (V)$ :
$T_{ma x} = \frac{1}{n R} (P_{0} \frac{P _{0}}{3 α} - α (\frac{P _{0}}{3 α})^{3})$
$T_{ma x} = \frac{1}{n R} (\frac{P _{0}^{3/2}}{3 α} - α \frac{P _{0}^{3/2}}{3 3 α ^{3/2}}) = \frac{P _{0}^{3/2}}{n R 3 α} (1 - \frac{1}{3})$
$T_{ma x} = \frac{2 P _{0}^{3/2}}{3 n R 3 α}$

This result shows how the maximum temperature depends on the parameters of the process, a typical query in advanced questions. This method of expressing a quantity as a function of one variable and using differentiation is a powerful and versatile tool.

4. Challenge Problem: When does Heat Flow Reverse?

Now for a problem that combines concepts. The following image from a JEE Advanced workbook presents a process involving a straight-line path, but the question asked is more subtle and requires a deeper analysis using the differential form of the first law.

Thermodynamics JEE Advanced Practice Problems — This image contains two problems. Let's focus on the first one, involving the process from A to B.

Problem Statement: One mole of a monoatomic gas is taken from A to B along the path shown.
(a) Find the temperature and volume at which heat supplied to the gas changes its sign.
(b) Find the heat supplied to the gas during the process A → B.

This is an excellent challenge. Give it a try before revealing the solution.

Test your understanding!

Hints for part (a):

"Heat supplied changes its sign" means we are looking for the point where $d Q = 0$ .
Start with the First Law in differential form: $d Q = d U + d W$ .
Express $d U$ and $d W$ in terms of a single variable, like $V$ , and its differential $d V$ . Remember $d U = n C_{v} d T$ and $d W = P d V$ .
You'll first need the equation for the line A→B, i.e., $P (V)$ .

Hints for part (b):

Use the integral form: $Q_{A \to B} = Δ U_{A \to B} + W_{A \to B}$ .
You've done this before! Calculate work from the area of the trapezoid and $Δ U$ from the endpoints.

Show solution

Part (b) - Total Heat Supplied (easier to calculate first):

Work Done $W_{A \to B}$ : Area of the trapezoid under A-B.
$W_{A \to B} = \frac{1}{2} (P_{A} + P_{B}) (V_{B} - V_{A}) = \frac{1}{2} (2 P_{0} + P_{0}) (2 V_{0} - V_{0}) = \frac{3}{2} P_{0} V_{0}$
Change in Internal Energy $Δ U_{A \to B}$ :
For one mole of a monoatomic gas, $C_{v} = \frac{3}{2} R$ .
$Δ U_{A \to B} = n C_{v} (T_{B} - T_{A}) = \frac{C _{v}}{R} (P_{B} V_{B} - P_{A} V_{A})$
Here, $P_{A} V_{A} = (2 P_{0}) (V_{0}) = 2 P_{0} V_{0}$ and $P_{B} V_{B} = (P_{0}) (2 V_{0}) = 2 P_{0} V_{0}$ .
Therefore, $Δ U_{A \to B} = \frac{3}{2 R} (2 P_{0} V_{0} - 2 P_{0} V_{0}) = 0$ .
This implies the process starts and ends on the same isotherm ( $T_{A} = T_{B}$ ).
Heat Supplied $Q_{A \to B}$ :
$Q_{A \to B} = Δ U_{A \to B} + W_{A \to B} = 0 + \frac{3}{2} P_{0} V_{0} = \frac{3}{2} P_{0} V_{0}$

Part (a) - Point of Heat Flow Reversal:

We are looking for the point where $d Q = 0$ .
$d Q = d U + P d V = n C_{v} d T + P d V = 0$
We need to relate $d T$ to $d V$ . First, find the equation of the line A $(V_{0}, 2 P_{0})$ to B $(2 V_{0}, P_{0})$ .
The slope is $m = \frac{P _{0} - 2 P _{0}}{2 V _{0} - V _{0}} = - \frac{P _{0}}{V _{0}}$ .
The equation is $P - P_{0} = - \frac{P _{0}}{V _{0}} (V - 2 V_{0}) ⟹ P (V) = - \frac{P _{0}}{V _{0}} V + 3 P_{0}$ .

Now express $T$ as a function of $V$ :
$T (V) = \frac{P ( V ) V}{n R} = \frac{1}{n R} (- \frac{P _{0}}{V _{0}} V^{2} + 3 P_{0} V)$
Differentiate to find $d T$ in terms of $d V$ :
$d T = \frac{1}{n R} (- \frac{2 P _{0}}{V _{0}} V + 3 P_{0}) d V$
Substitute this into the $d Q = 0$ equation:
$n C_{v} [\frac{1}{n R} (- \frac{2 P _{0}}{V _{0}} V + 3 P_{0}) d V] + (- \frac{P _{0}}{V _{0}} V + 3 P_{0}) d V = 0$
Divide by $d V$ and substitute $C_{v} / R = 3/2$ for a monoatomic gas:
$\frac{3}{2} (- \frac{2 P _{0}}{V _{0}} V + 3 P_{0}) + (- \frac{P _{0}}{V _{0}} V + 3 P_{0}) = 0$
$(- \frac{3 P _{0}}{V _{0}} V + \frac{9}{2} P_{0}) + (- \frac{P _{0}}{V _{0}} V + 3 P_{0}) = 0$
$- \frac{4 P _{0}}{V _{0}} V + \frac{15}{2} P_{0} = 0 ⟹ V = \frac{15}{8} V_{0}$
Now find the temperature at this volume:
$P = - \frac{P _{0}}{V _{0}} (\frac{15}{8} V_{0}) + 3 P_{0} = (- \frac{15}{8} + \frac{24}{8}) P_{0} = \frac{9}{8} P_{0}$
$T = \frac{P V}{n R} = \frac{( \frac{9}{8} P _{0} ) ( \frac{15}{8} V _{0} )}{n R} = \frac{135 P _{0} V _{0}}{64 n R}$
So, heat flow reverses at $V = 1.875 V_{0}$ . Initially, as temperature rises (you can check that $T_{ma x}$ occurs at $V = 1.5 V_{0}$ ), heat is absorbed. After $V = 1.875 V_{0}$ , the gas is still expanding (doing positive work), but it's cooling down fast enough that it must reject heat to follow the path.

Conclusion

This lesson completes your toolkit for analyzing any thermodynamically defined process path you might encounter in the JEE. By moving from standard processes to linear paths and now to general curved paths, you've seen that the core principles remain the same, with calculus serving as the universal problem-solving engine.

Key Takeaways:

For any process defined by a function $P (V)$ , work done is found by direct integration: $W = \int P (V) d V$ .
Changes in state variables ( $Δ P, Δ V, Δ T$ ) are found by applying the process equation and the ideal gas law at the initial and final points.
The change in internal energy, $Δ U$ , always depends only on the change in temperature, $Δ T$ , regardless of the path's complexity.
Optimization techniques (setting a derivative to zero) are powerful for finding maximum or minimum values of state variables like temperature during a process.
The differential form of the first law, $d Q = d U + d W$ , is essential for analyzing instantaneous changes, such as finding where heat flow reverses.

Preview of the Next Lesson:
You now have the tools to deconstruct any individual process, no matter how unconventional. In our next lesson, "Apply cycle analysis to solve complex JEE Advanced problems with unconventional thermodynamic cycles," we will focus on synthesis. You will learn to combine these techniques to analyze full, multi-stage cycles composed of these complex paths, calculating net work, net heat, and overall efficiency—the pinnacle of thermodynamic problem-solving for the exam.

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