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

From the given graph, the y-axis represents volume ($V$) and the x-axis represents pressure ($P$).

The total work done in the process $a \rightarrow b \rightarrow c$ is the sum of the work done in processes $a \rightarrow b$ and $b \rightarrow c$.

For the process $a \rightarrow b$, the volume remains constant at $V = 100 \text{ cm}^3$. Since it is an isochoric process, the change in volume $dV = 0$.

Work done $W_{ab} = \int P \, dV = 0$.

For the process $b \rightarrow c$, the pressure remains constant at $P = 200 \text{ kPa}$. Since it is an isobaric process, the work done is given by $W_{bc} = P(V_c - V_b)$.

Substituting the given values:
$P = 200 \text{ kPa} = 200 \times 10^3 \text{ Pa}$
$V_b = 100 \text{ cm}^3 = 100 \times 10^{-6} \text{ m}^3$
$V_c = 150 \text{ cm}^3 = 150 \times 10^{-6} \text{ m}^3$

$W_{bc} = (200 \times 10^3) \times (150 - 100) \times 10^{-6}$
$W_{bc} = 200 \times 10^3 \times 50 \times 10^{-6}$
$W_{bc} = 10000 \times 10^{-3} = 10 \text{ J}$

Total work done $W = W_{ab} + W_{bc} = 0 + 10 = 10 \text{ J}$.

Answer: $10 \text{ J}$