Q. A process is taking place at constant temperature and pressure. Then
A) $\mathrm{\Delta}H = \mathrm{\Delta}E$
B) $\mathrm{\Delta}H = T\mathrm{\Delta}S$
C) $\mathrm{\Delta}H = 0$
D) $\mathrm{\Delta}S = 0$
Explanation
(a)
For an isothermal process, $\mathrm{\Delta}E = 0$
As the process is taking place at constant $T$ and $p$ hence, from equation,
$\mathrm{\Delta}H = \mathrm{\Delta}E + \mathrm{\Delta}p.V$
We have, $\mathrm{\Delta}H = 0 + 0 \times V = 0$
Hence, for the process, $\mathrm{\Delta}H = \mathrm{\Delta}E = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.