Chapter 20 - Entropy and the Second Law of Thermodynamics - Problems - Page 608: 62h

$\Delta E_{int} = 1247.1 J = 1.25 kJ $

The change in internal energy for adiabatic process in step $2 \rightarrow 3 $ is in the following with $C_V = \frac{3}{2} $ and $n = 2.00 mol$ $\Delta E_{int} = nC_V (T_3 - T_2)$ $\Delta E_{int} = (2.00 mol)(\frac{3}{2} \times 8.314 J/k.mol) (350 K - 300K)$ $\Delta E_{int} = 1247.1 J = 1.25 kJ $