PYQ + Trends
1. TITLE
NEET 2020 Q16: Adiabatic Processes & Free Expansion — Full solution + Why they asked this
2. THE ACTUAL NEET QUESTION
NEET 2020 — Question
The correct option for free expansion of an ideal gas under adiabatic condition is:
A) $q=0$, $\Delta T<0$ and $w>0$
B) $q<0$, $\Delta T=0$ and $w=0$
C) $q>0$, $\Delta T>0$ and $w>0$
D) $q=0$, $\Delta T=0$ and $w=0$
3. OFFICIAL ANSWER
NEET Answer Key: (D)
4. COMPLETE SOLUTION
For Physical Chemistry:
- Step-by-step calculation:
1. "Adiabatic condition": The system is perfectly insulated, meaning there is no exchange of heat between the system and the surroundings. Therefore, $q = 0$.
2. "Free expansion": The gas is expanding into a vacuum. In a vacuum, the external pressure opposing the expansion is zero ($P_{ex} = 0$). Since work done is calculated as $w = -P_{ex}\Delta V$, setting $P_{ex}$ to zero means the work done is also zero ($w = 0$).
3. First Law of Thermodynamics: The law states that $\Delta U = q + w$. Since both $q = 0$ and $w = 0$, the change in internal energy is zero ($\Delta U = 0$).
4. Temperature Change: For an ideal gas, internal energy depends exclusively on temperature ($\Delta U = nC_v\Delta T$). Since $\Delta U = 0$, it directly implies that the change in temperature is zero ($\Delta T = 0$).
- Formula used:
- $w = -P_{ex}\Delta V$
- $\Delta U = q + w$
- $\Delta U = nC_v\Delta T$
- Why this approach: By systematically breaking down the two keywords—"adiabatic" and "free expansion"—you immediately find the values of $q$ and $w$, which unlock $\Delta U$ and subsequently $\Delta T$ via the First Law of Thermodynamics.
5. WHY NEET ASKED THIS
Explain the examiner's intent:
- What concept were they testing? They were testing the mathematical implications of specific thermodynamic constraints—specifically, that "vacuum" removes the work component and "adiabatic" removes the heat component.
- What's the common mistake they expected students to make? Students frequently memorize that "adiabatic expansion causes cooling". While normal adiabatic expansion against atmospheric pressure does do work at the expense of internal energy ($\Delta U = w$), causing a temperature drop ($\Delta T < 0$), "free expansion" is the exception where no work is done.
- Which option was the "trap"? Option A ($q=0$, $\Delta T<0$ and $w>0$) and similar variations were the traps designed to catch students who associate all expansions with temperature drops, ignoring the $P_{ex}=0$ rule.
6. TREND ANALYSIS
PATTERN ALERT:
| Year | How they asked | Twist |
|---|---|---|
| 2024 | Match the following process to conditions. | Basic conceptual check. "Adiabatic process" directly matched with "No heat exchange" ($q=0$). |
| 2020 | Conceptual conditions for "Free expansion + Adiabatic". | Tested 3 variables ($q$, $w$, $\Delta T$) simultaneously. |
| 2017 | Calculate internal energy ($\Delta U$) for an insulated container expanding against 2.5 atm. | Applied calculation. Because it was not free expansion ($P_{ex} = 2.5$), work was done, so $\Delta U = w = -506.625$ J. |
| 2011 | Exact same phrasing: "Correct option for free expansion of an ideal gas under adiabatic condition". | Zero twist. NEET 2020 recycled the exact conceptual PYQ from 2011 verbatim! |
Prediction for NEET 2026:
Expect a graph-based variation or calculation. You may be asked to compare the final temperature or work done between an isothermal expansion and an adiabatic expansion on a P-V graph, where adiabatic curves are steeper and involve a drop in temperature.
7. SIMILAR QUESTIONS TO PRACTICE
"If you understood this, you can solve:"
- NEET 2011 Q40: The original mirror-image PYQ for free adiabatic expansion.
- NEET 2017 Q24: Adiabatic expansion against constant external pressure to calculate exact Joules of internal energy change.
- NEET 2024 Q7: Direct match-the-following regarding isochoric, isobaric, isothermal, and adiabatic boundary conditions.
8. QUICK REVISION FOR THIS TYPE
For Adiabatic & Free Expansion questions, always:
- First check: Is it expanding into a vacuum ("free expansion")? If yes, $P_{ex} = 0$, so $w = 0$.
- Then apply: Is the container "insulated" or "adiabatic"? If yes, heat transfer is blocked, so $q = 0$.
- Watch out for: The First Law formula ($\Delta U = q + w$). If both $q$ and $w$ are zero, $\Delta U = 0$. Since ideal gas internal energy is temperature-dependent, the temperature must stay completely constant ($\Delta T = 0$). Don't assume all adiabatic expansions cause a temperature drop!
9. NCERT BASE
This question tests: NCERT Class 11 Chemistry Part 1, Chapter 5 (Thermodynamics), Page 142 (Under Section 5.2.1: Free expansion and Isothermal/Adiabatic processes).
10. CTA
All PYQs from this chapter → Chemical Thermodynamics Guide
Practice similar questions → Mock exams