Answer
$Q_H =25.5 kJ $
Work Step by Step
To find the net energy transferred, $Q_H$. we need to know the processes involved. Note that only process 1 and 2 are involved in acquiring heat as energy.
Process 1 : Isochoric, $dV = 0, W = 0 , dU = dQ$
Process 2 : Isothermal, $dT = 0, Q = W$
$Q_H = Q_1 + Q_2 $
$Q_H = nC_V (T' - T) + nRT' ln (\frac{T'}{T})$
$Q_H = n\frac{3}{2}R (T' - T) + nRT' ln (\frac{T'}{T})$
$Q_H = (2.0 mol) \frac{3}{2}(8.31 J/mol.K) (800K' - 300K) + (2.0 mol) (8.31 J/mol.K)(800K) ln (\frac{800K'}{300K})$
$Q_H = 12465 J + 13041 J$
$Q_H = 25.5 \times 10^3 J = 25.5 kJ$
