An ideal gas ( γ = 1.67 ) is taken through the process abc shown in...

From the given graph, the pressure and volume at the initial state $a$ and final state $c$ are:

$P_a = 100 \text{ kPa} = 10^5 \text{ Pa}$
$V_a = 100 \text{ cm}^3 = 10^{-4} \text{ m}^3$
$P_c = 200 \text{ kPa} = 2 \times 10^5 \text{ Pa}$
$V_c = 150 \text{ cm}^3 = 1.5 \times 10^{-4} \text{ m}^3$

The product of pressure and volume at these states can be calculated as:

$P_a V_a = 10^5 \times 10^{-4} = 10 \text{ J}$
$P_c V_c = 2 \times 10^5 \times 1.5 \times 10^{-4} = 30 \text{ J}$

The change in internal energy $\Delta U$ for an ideal gas depends only on the initial and final states and is given by:

$\Delta U = \dfrac{P_c V_c - P_a V_a}{\gamma - 1}$

Substituting the given values ($\gamma = 1.67$):

$\Delta U = \dfrac{30 - 10}{1.67 - 1}$

$\Delta U = \dfrac{20}{0.67} \approx 29.85 \text{ J}$

The closest matching option is $29.8 \text{ J}$.

Answer: $29.8 \text{ J}$