Q. In a thermodynamic process pressure of a fixed mass of a gas is changed in such a manner that the gas releases $30\ joules$ of heat and 10 joules of work was done on the gas. If the initial internal energy of the gas was $30\ joules$, then the final internal energy will be
A) $2\ J$
B) $- 18\ J$
C) $10\ J$
D) $58\ J$
Explanation
(c) $\mathrm{\Delta}Q = \mathrm{\Delta}U + \mathrm{\Delta}W = \left( U_{f} - U_{i} \right) + \mathrm{\Delta}W$
$\Rightarrow - 30 = \left( U_{f} - 30 \right) - 10 \Rightarrow U_{f} = 10\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.