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JEE Advanced 2025Paper 1
Exam
JEE Advanced
Year
2025
Paper
Paper 1
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Advanced 2025 · Paper 1 · Chemistry · Thermodynamics (C)

Considering ideal gas behavior, the expansion work done (in kJ)...

Considering ideal gas behavior, the expansion work done (in kJ) when 144 g of water is electrolyzed completely under constant pressure at 300 K is _____ Use: Universal gas constant (R)=8.3 J K−1 mol−1(\mathrm{R})=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}; Atomic mass (in amu): H=1,O=16\mathrm{H}=1, \mathrm{O}=16 (mark absolute value as answer)

Correct answer

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Step-by-step solution

in official answer key answer was [-29.95 to -29.8] OR [29.8 to 29.95], we have updated the question to make it valid. H2O(l)→H2(g)+12O2(g)144g−−8mol8mol4molW=−PΔV=−(Δn)RT\begin{array}{lcc} \mathrm{H}_2 \mathrm{O}(l) \rightarrow & \mathrm{H}_2(\mathrm{ g})+ & \frac{1}{2} \mathrm{O}_2(\mathrm{ g}) \\ 144 \mathrm{ g} & - & - \\ 8 \mathrm{ mol} & 8 \mathrm{ mol} & 4 \mathrm{ mol} \\ \mathrm{ W}=-\mathrm{P} \Delta \mathrm{ V}=& -(\Delta \mathrm{n}) \mathrm{RT} \end{array} Change in gaseous moles =12⇒W=−12×8.3×3001000kJ=−29.88kJ.\begin{aligned} &\text{ Change in gaseous moles }=12\\ &\Rightarrow \mathrm{W}=-\frac{12 \times 8.3 \times 300}{1000} \mathrm{ kJ}=-29.88 \mathrm{ kJ} . \end{aligned}$
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