Chapter 20 - Entropy and the Second Law of Thermodynamics - Problems - Page 605: 19e

$\frac{\Delta E_{int}}{ p_1V_1 } = 4.50$

The change of internal energy is given by $\Delta E_{int} = Q_{total} - W_{total}$ $\Delta E_{int} = (p_1V_1 ln 2 + \frac{9}{2} p_1V_1) - (p_1V_1 ln 2 + 0)$ $\Delta E_{int} = \frac{9}{2} p_1V_1 $ $\frac{\Delta E_{int}}{ p_1V_1 } = \frac{9}{2}$ $\frac{\Delta E_{int}}{ p_1V_1 } = 4.50$