Answer to Question #252154 in Molecular Physics | Thermodynamics for elakkiya

Question

A closed system undergoes a reversible process at a constant pressure process of 3.5 bar and its volume changes from 0.15 m ³to 06m ³.25 kJ of heat is rejected by the system during the process. Determine the change in internal energy of the system.

Answer 1

$\Delta U=Q-A=Q-p\Delta V=Q-p(V_2-V_1),$

$\Delta U=25000-350000\cdot(0.06-0.15)=56.5~\text{kJ}.$