Q. We consider a thermodynamic system. If $\mathrm{\Delta}U$ represents the increase in its internal energy and $W$ the work done by the system, which of the following statements is true?
A) $\mathrm{\Delta}U = - W$ in an adiabatic process
B) $\mathrm{\Delta}U = W$in an isothermal process
C) $\mathrm{\Delta}U = - W$ in an isothermal process
D) $\mathrm{\Delta}U = W$in an adiabatic process
Explanation
(a) An isothermal process is a constant temperature process. In this process, $T = constant\ or\ \mathrm{\Delta}T = 0.$
$\ \therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathrm{\Delta}U = nC_{V}\ \mathrm{\Delta}T = 0$
An adiabatic process is defined as one with no heat transfer into or out of a system. Therefore, $\mathrm{\Delta}Q = 0.$ From the first law of thermodynamics.
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ W = - \mathrm{\Delta}U$
or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathrm{\Delta}U = - W$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.