Q. The work done during the expansion of a gas from a volume of $4\ dm^{3}\ to\ 6\ {dm}^{3}$ against a constant external pressure of 3 atm is
A) -6 J
B) -608 J
C) +304 J
D) -304 J
Explanation
(b)
Work done(W)=$- p_{ext}(V_{2} - V_{1})$
$= - 3 \times (6 - 4) = - 6\ L\ atm$
$\ = \ - 6 \times 101.32$ J($\therefore$ 1 L atm=101.32 J)
$= \ - 607.92 \approx - 608\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.