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

From the given graph:

Path $a \rightarrow b$ is an isochoric process (constant volume) where $V = 100 \text{ cm}^3 = 10^{-4} \text{ m}^3$.
The pressure changes from $P_a = 100 \text{ kPa} = 10^5 \text{ Pa}$ to $P_b = 200 \text{ kPa} = 2 \times 10^5 \text{ Pa}$.
The heat supplied at constant volume is given by:
$Q_{ab} = nC_V\Delta T = \dfrac{V\Delta P}{\gamma - 1}$
Substituting the values:
$Q_{ab} = \dfrac{10^{-4} \times (2 \times 10^5 - 10^5)}{1.67 - 1}$
$Q_{ab} = \dfrac{10^{-4} \times 10^5}{0.67} = \dfrac{10}{0.67} \approx 14.9 \text{ J}$

Path $b \rightarrow c$ is an isobaric process (constant pressure) where $P = 200 \text{ kPa} = 2 \times 10^5 \text{ Pa}$.
The volume changes from $V_b = 100 \text{ cm}^3 = 10^{-4} \text{ m}^3$ to $V_c = 150 \text{ cm}^3 = 1.5 \times 10^{-4} \text{ m}^3$.
The heat supplied at constant pressure is given by:
$Q_{bc} = nC_P\Delta T = \dfrac{\gamma P\Delta V}{\gamma - 1}$
Substituting the values:
$Q_{bc} = \dfrac{1.67 \times 2 \times 10^5 \times (1.5 \times 10^{-4} - 10^{-4})}{1.67 - 1}$
$Q_{bc} = \dfrac{1.67 \times 2 \times 10^5 \times 0.5 \times 10^{-4}}{0.67}$
$Q_{bc} = \dfrac{1.67 \times 10}{0.67} = \dfrac{16.7}{0.67} \approx 24.9 \text{ J}$

The heat supplied in paths $ab$ and $bc$ are $14.9 \text{ J}$ and $24.9 \text{ J}$ respectively.

Answer: $14.9 \text{ J}$, $24.9 \text{ J}$