Half mole of an ideal gas ( γ = 5/3 ) is taken through a cycle...

Given for the ideal gas, $\gamma = \dfrac{5}{3}$.

The molar heat capacities are:
$C_v = \dfrac{R}{\gamma - 1} = \dfrac{R}{\dfrac{5}{3} - 1} = \dfrac{3}{2}R$
$C_p = \dfrac{\gamma R}{\gamma - 1} = \dfrac{5}{2}R$

From the $P$-$V$ diagram, the states are:
$P_a = 100 \text{ kPa} = 10^5 \text{ Pa}$, $V_a = 5000 \text{ cm}^3 = 5 \times 10^{-3} \text{ m}^3$
$P_b = 100 \text{ kPa} = 10^5 \text{ Pa}$, $V_b = 10000 \text{ cm}^3 = 10 \times 10^{-3} \text{ m}^3$
$P_c = 200 \text{ kPa} = 2 \times 10^5 \text{ Pa}$, $V_c = 10000 \text{ cm}^3 = 10 \times 10^{-3} \text{ m}^3$

Process $ab$ is isobaric (constant pressure). The heat supplied is:
$Q_{ab} = nC_p\Delta T = \dfrac{C_p}{R}(nR\Delta T) = \dfrac{C_p}{R}P\Delta V$
$Q_{ab} = \dfrac{5}{2} P_a (V_b - V_a)$
$Q_{ab} = \dfrac{5}{2} \times 10^5 \times (10 \times 10^{-3} - 5 \times 10^{-3})$
$Q_{ab} = \dfrac{5}{2} \times 10^5 \times 5 \times 10^{-3} = \dfrac{5}{2} \times 500 = 1250 \text{ J}$

Process $bc$ is isochoric (constant volume). The heat supplied is:
$Q_{bc} = nC_v\Delta T = \dfrac{C_v}{R}(nR\Delta T) = \dfrac{C_v}{R}V\Delta P$
$Q_{bc} = \dfrac{3}{2} V_b (P_c - P_b)$
$Q_{bc} = \dfrac{3}{2} \times 10 \times 10^{-3} \times (2 \times 10^5 - 10^5)$
$Q_{bc} = \dfrac{3}{2} \times 10^{-2} \times 10^5 = \dfrac{3}{2} \times 1000 = 1500 \text{ J}$

Therefore, the heat supplied in processes $ab$ and $bc$ are $1250 \text{ J}$ and $1500 \text{ J}$ respectively.

Answer: $1250 \text{ J}$, $1500 \text{ J}$