Numerical/Formula
1. TITLE
How to solve Reversible Isothermal Work numericals in NEET — Step-by-step
2. PROBLEM TYPE INTRO
This category of numerical tests your ability to calculate the maximum work done during the isothermal reversible expansion or compression of an ideal gas. We will use the integrated work equation: $w_{rev} = -2.303 nRT \log(V_f/V_i)$. The typical mistake students make is messing up the IUPAC sign convention (expansion is negative work) or forgetting to multiply by the $2.303$ conversion factor for base-10 logarithms.
3. MCQ
Calculate the maximum work done when $2 \text{ moles}$ of an ideal gas expand isothermally and reversibly from a volume of $10 \text{ L}$ to $100 \text{ L}$ at $27^\circ\text{C}$. ($R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$)
(a) $-11.49 \text{ kJ}$
(b) $+11.49 \text{ kJ}$
(c) $-4.99 \text{ kJ}$
(d) $-114.9 \text{ J}$
4. STEP-BY-STEP SOLUTION
SOLUTION
Given:
* Number of moles ($n$) = $2 \text{ mol}$
* Temperature ($T$) = $27^\circ\text{C}$
* Initial volume ($V_i$) = $10 \text{ L}$
* Final volume ($V_f$) = $100 \text{ L}$
* Universal gas constant ($R$) = $8.314 \text{ J K}^{-1} \text{ mol}^{-1}$
To Find:
* The reversible work done ($w_{rev}$) in kilojoules (kJ).
Formula:
* $W_{rev} = -2.303 nRT \log(V_f/V_i)$
* Explain when to use this formula: Use this formula specifically when the problem states the process is isothermal (constant temperature) and reversible (proceeding infinitely slowly through equilibrium states) for an ideal gas.
Step 1: Convert temperature to the absolute scale (Kelvin).
* $T = 27 + 273 = 300 \text{ K}$
Step 2: Substitute the values into the reversible work formula.
* $w_{rev} = -2.303 \times 2 \text{ mol} \times 8.314 \text{ J K}^{-1} \text{ mol}^{-1} \times 300 \text{ K} \times \log(100/10)$
Step 3: Solve the logarithmic part and calculate the final value.
* $\log(100/10) = \log(10) = 1$
* $w_{rev} = -2.303 \times 2 \times 8.314 \times 300 \times 1$
* $w_{rev} = -11488.28 \text{ J}$
* Convert Joules to kilojoules by dividing by $1000$: $-11.488 \text{ kJ}$
Answer: $-11.49 \text{ kJ}$ (Option a)
5. WHERE STUDENTS GO WRONG
List 2-3 common mistakes:
* ❌ Mistake 1: Using the wrong sign for work. In IUPAC convention, work done by the system during expansion decreases internal energy, so it must be negative (Option b result).
* ❌ Mistake 2: Forgetting the $2.303$ factor. Students often use the $\ln$ (natural log) formula but plug in base-10 log values, skipping the $2.303$ multiplier, resulting in $-4.99 \text{ kJ}$ (Option c result).
* ❌ Mistake 3: Failing to convert units. Not converting Celsius to Kelvin, or forgetting to divide the final Joules by $1000$ to match the kilojoule options (Option d result).
6. QUICK FORMULA REFERENCE
Formulas Used:
* $W_{rev} = -2.303 nRT \log(V_f/V_i)$ — Use when calculating work for an isothermal reversible expansion/compression of an ideal gas using volume changes.
* $w = -p_{ex} \Delta V$ — Use when calculating work for an isothermal irreversible process against a constant external pressure.
7. SIMILAR PROBLEM HINT
"If NEET changes the values, the method stays the same. Practice with: Reversible isothermal work where initial and final PRESSURES are given instead of volumes. (Hint: Use Boyle's Law $V_f/V_i = p_i/p_f$ to adapt the formula to $w = -2.303 nRT \log(p_i/p_f)$)."
8. NCERT REFERENCE
Based on: NCERT Class 11, Chapter 5 (Thermodynamics), Problem 5.4.
9. CTA
More solved numericals → Chemical Thermodynamics Complete Guide
Practice problems → Mock exams