Q. In changing the state of a gas adiabatically from an equilibrium state $A$ to another equilibrium state $B$, an amount of work equal to $22.3\ J$ is done on the system. If the gas is taken from state $A$ to $B$ via a process in which the net heat absorbed by the system is 9.35 calories, the net work done by the system in latter case will be
A) $5.9\ J$
B) $16.9\ J$
C) $9.3\ J$
D) $4.6\ J$
Explanation
(b) According to $I^{st}$ law of Thermodynamics
$\mathrm{\Delta}Q = \mathrm{\Delta}U + \mathrm{\Delta}W$, in adiabatic process $\mathrm{\Delta}Q = 0$
$0 = \mathrm{\Delta}U - \mathrm{\Delta}W$ (Work done on the system –Ve)
$\mathrm{\Delta}U = + \mathrm{\Delta}W = + 22.3$
[ $\therefore$ Work done on the system $\therefore$ internal energy increases]
In $2^{nd}$ process $\mathrm{\Delta}Q = \mathrm{\Delta}U + \mathrm{\Delta}W$
$9.35 \times 4.18 = 22.3 + \mathrm{\Delta}W$
Work done by system $\mathrm{\Delta}W = 16.95$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.