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

Given $\gamma = \dfrac{5}{3}$, the molar heat capacities are $C_v = \dfrac{R}{\gamma - 1} = \dfrac{3}{2}R$ and $C_p = C_v + R = \dfrac{5}{2}R$.

For the isobaric process $cd$, the pressure is constant at $P = 200 \text{ kPa} = 2 \times 10^5 \text{ Pa}$.

The heat exchanged is $Q_{cd} = n C_p \Delta T = \dfrac{C_p}{R} P \Delta V = \dfrac{5}{2} P (V_d - V_c)$.

Substituting the given values:

$Q_{cd} = \dfrac{5}{2} \times (2 \times 10^5) \times (5000 - 10000) \times 10^{-6} = -2500 \text{ J}$.

The negative sign indicates that $2500 \text{ J}$ of heat is liberated.

For the isochoric process $da$, the volume is constant at $V = 5000 \text{ cm}^3 = 5 \times 10^{-3} \text{ m}^3$.

The heat exchanged is $Q_{da} = n C_v \Delta T = \dfrac{C_v}{R} V \Delta P = \dfrac{3}{2} V (P_a - P_d)$.

Substituting the given values:

$Q_{da} = \dfrac{3}{2} \times (5 \times 10^{-3}) \times (100 - 200) \times 10^3 = -750 \text{ J}$.

The negative sign indicates that $750 \text{ J}$ of heat is liberated.

Thus, the heat liberated in processes $cd$ and $da$ are $2500 \text{ J}$ and $750 \text{ J}$ respectively.

Answer: $2500 \text{ J}$, $750 \text{ J}$