PYQ + Trends
1. TITLE
NEET asked Work Done in Isothermal Expansion in 2019, 2024 — Expect it in 2025
2. THE ACTUAL NEET QUESTION
NEET 2019 — Question
Under isothermal conditions, a gas at 300 K expands from 0.1 L to 0.25 L against a constant external pressure of 2 bar. The work done by the gas is [Given that 1L bar = 100J ]
A) 30J
B) -30J
C) 5kJ
D) 25J
3. OFFICIAL ANSWER
NEET Answer Key: (B)
4. COMPLETE SOLUTION
For Physics/Physical Chemistry:
- Step-by-step calculation:
1. Identify the given values: Initial volume ($V_1$) = 0.1 L, Final volume ($V_2$) = 0.25 L, Constant external pressure ($P_{ex}$) = 2 bar.
2. Calculate the change in volume ($\Delta V$): $V_2 - V_1 = 0.25 \text{ L} - 0.1 \text{ L} = 0.15 \text{ L}$.
3. Apply the work formula: $w = -P_{ex} \Delta V = -2 \text{ bar} \times 0.15 \text{ L} = -0.30 \text{ L bar}$.
4. Convert units to Joules using the given conversion factor (1 L bar = 100 J): $w = -0.30 \times 100 \text{ J} = -30 \text{ J}$.
- Formula used: $w = -p_{ex}\Delta V = -p_{ex}(V_f - V_i)$.
- Why this approach: Expansion of a gas against a constant external pressure in a single step is an irreversible process, which dictates the use of the basic pressure-volume work equation ($w = -P_{ex}\Delta V$) rather than the logarithmic formula used for reversible processes.
5. WHY NEET ASKED THIS
Explain the examiner's intent:
- What concept were they testing? The examiner was testing two things: the application of the irreversible work formula and the IUPAC sign convention for thermodynamic work.
- What's the common mistake they expected students to make? Forgetting that work done by the system (expansion) leads to a decrease in internal energy and must have a negative sign. Students often calculate the magnitude (30 J) and ignore the negative sign.
- Which option was the "trap"? Option A (30J) is the primary trap, sitting right at the top for students who miss the negative sign in the expansion formula.
6. TREND ANALYSIS
PATTERN ALERT:
| Year | How they asked | Twist |
|---|---|---|
| 2024 (Re) | Asked for the work done and heat absorbed when 5 litres of an ideal gas at 10 atm isothermally expands into a vacuum until volume is 15 litres. | The external pressure is zero ($P_{ex} = 0$). This is "free expansion," meaning work done ($w$) is zero, and since it is isothermal, heat ($q$) is also zero! |
| 2019 | Gas expands from 0.1 L to 0.25 L against a constant external pressure of 2 bar. | Direct calculation with a specific unit conversion trick given in the prompt (1 L bar = 100 J). |
| 2019 (Odisha) | An ideal gas expands isothermally from $10^{-3} m^3$ to $10^{-2} m^3$ at 300K against a constant pressure of $10^5 N m^{-2}$. | Similar to the main 2019 paper, but utilizing SI units ($m^3$ and $N m^{-2}$) requiring knowledge that $1 J = 1 N m$. |
Prediction for NEET 2025:
Expect a variation where the gas is compressed (making work positive) at constant pressure, or a conceptual hybrid question asking you to identify which process (reversible vs. irreversible vs. free expansion into vacuum) yields maximum work.
7. SIMILAR QUESTIONS TO PRACTICE
"If you understood this, you can solve:"
- NEET 2024 Re Q3: Isothermal expansion into a vacuum.
- Odisha NEET 2019 Q20: Expansion against constant $10^5 N m^{-2}$ pressure.
- NEET 2024 Q8: Reversible isothermal expansion of hydrogen gas.
8. QUICK REVISION FOR THIS TYPE
For Work Done in Isothermal Expansion ($w=-P_{ex}\Delta V$) questions, always:
- First check: Is the process expanding into a vacuum? If it says "free expansion" or "vacuum", $P_{ex} = 0$, so $w = 0$ instantly.
- Then apply: If a constant external pressure is given, use $w = -p_{ex}(V_f - V_i)$. If it is a reversible process, use $w_{rev} = -2.303 nRT \log(V_f/V_i)$.
- Watch out for: Sign conventions and unit conversions! Expansion means final volume is greater than initial volume, making $\Delta V$ positive, which makes overall work negative (work done by the system). Always double-check if you need to multiply by 101.3 (for L-atm to Joules) or 100 (for L-bar to Joules).
9. NCERT BASE
This question tests: NCERT Class 11, Chapter 5 (Thermodynamics), Page 141.
10. CTA
All PYQs from this chapter → Chemical Thermodynamics Guide
Practice similar questions → Mock exams