Q. A thermodynamical system goes from state (i) $(p,\ V)$ to $(2\ p,\ V)$ and (ii) $(p,\ V)$ to $(p,\ 2V)$. Work done in the two cases is
A) Zero, zero
B) Zero, $pV$
C) $pV$, zero
D) $pV,\ pV$
Explanation
(b) As $dW = p\ dV$
$\therefore$ (i) $dW = p \times 0 = 0$
(ii) $dW = p(2V - V) = pV$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.