Problem 17.19

17.19.
Work done is given by the expression
$W_{ab} = \int_{a}^{b}P dV$
Here, if we plug in out expression for the pressure,
$W_{ab} = \int_{a}^{b} \alpha V^{2} dV$
$W_{ab} = \frac{1}{3} \alpha (V_{b}^{3} - V_{a}^{3})$
we know from the graph that V_a = 1 m³ and V_b = 2 m³.
So, the work done is
$W_{ab} = \frac{1}{3} (5.0 atm/m^{6})(101.3 kPa/atm)[(2^{3} - 1^{3}) m^{9}]$
$W_{ab} = 1.18 \times 10^{6} J$