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JEE Main 20253 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
3 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Main 2025 · 3 April · Evening Shift · Chemistry · Thermodynamics (C)

A perfect gas (0.1 ~mol) having C _ v =1.50 R (independent of...

A perfect gas (0.1 ~mol) having C _ v =1.50 R (independent of temperature) undergoes the above transformation from
A perfect gas (0.1 mol)(0.1 \mathrm{~mol}) having C‾v=1.50R\overline{\mathrm{C}}_{\mathrm{v}}=1.50 \mathrm{R} (independent of temperature) undergoes the above transformation from point 1 to point 4. If each step is reversible, the total work done (w) while going from point 1 to point 4 is (−)(-) ________ J (nearest integer) \left[\right.Given:R=0.082 L atm K−1 mol−1]Given : \left.\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right]

Correct answer

0

Step-by-step solution

W1→2=0&W2→3=−PΔV&=−3[2−1]&=−3atm−ℓ\begin{aligned}& \mathrm{W}_{1 \rightarrow 2}=0 \& \mathrm{ W}_{2 \rightarrow 3}=-\mathrm{P} \Delta \mathrm{ V} \& =-3[2-1] \& =-3 \mathrm{ atm}-\ell\end{aligned}\mathrm{W}_{3 \rightarrow 4}=0TotalworkdoneTotal work done=−3atm−ℓ&=−3×101.3 Joule &=−304 Joule \begin{aligned}& =-3 \mathrm{ atm}-\ell \& =-3 \times 101.3 \text{ Joule } \& =-304 \text{ Joule }\end{aligned}$
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