Q. In a thermodynamic process, pressure of a fixed mass of a gas is changed in such a manner that the gas molecules gives out $20\ J$ of heat and $10\ J$ of work is done on the gas. If the initial internal energy of the gas was $40\ J$, then the final internal energy will be
A) $30\ J$
B) $20\ J$
C) $60\ J$
D) $40\ J$
Explanation
(a) $\mathrm{\Delta}Q = - 20J;\ \mathrm{\Delta}W = - 10J$
$\mathrm{\Delta}Q = \left( U_{f} - U_{i} \right) + \mathrm{\Delta}W$
$\Rightarrow \ - 20 = \left( U_{f} - 40 \right) - 10 \Rightarrow U_{f} = - 10 + 40 = 30\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.