Answer
$\Phi_2-\Phi_1=6980\ kJ$
Work Step by Step
For an ideal gas:
$s_2-s_1=c_v\ln{\dfrac{T_2}{T_1}}+R\ln{\dfrac{v_2}{v_1}}$
Given $c_v=3.1156\ kJ/kg.K,\ T_2=353\ K,\ T_1=288\ K,$
$v_2=0.5\ m^3/kg,\ v_1=3\ m^3/kg,\ R=2.0769\ kJ/kg.K$:
$s_2-s_1=-3.087\ kJ/kg.K$
The increase in useful potential energy is then:
$\Phi_2-\Phi_1=-m[(u_1-u_2)-T_0(s_1-s_2)+P_0(v_1-v_2)]$
Given $m=8\ kg,\ u_1-u_2=c_v(T_1-T_2),\ T_0=298\ K,\ P_0=100\ kPa$:
$\Phi_2-\Phi_1=6980\ kJ$
