Identify the processes from the $V$-$T$ diagram

The cycle consists of alternating adiabatic and isobaric processes:

• $W \to X$ : adiabatic
• $X \to Y$ : isobaric
• $Y \to Z$ : adiabatic
• $Z \to W$ : isobaric

We need the heat absorbed along the path $X \to Y$.

Use adiabatic relation in $V$-$T$ form

For an ideal gas in a quasi-static adiabatic process, $T V^{\gamma-1} = \text{constant}$ For a monatomic gas, $\gamma = \frac{5}{3} \quad \Rightarrow \quad \gamma-1 = \frac{2}{3}$

Hence for $W \to X$, $T_W V_W^{2/3} = T_X V_X^{2/3}$ so $T_X = T_W \left(\frac{V_W}{V_X}\right)^{2/3}$

Given: $V_W = 64\ \text{cm}^3, \quad V_X = 125\ \text{cm}^3$

Therefore, $T_X = T_W \left(\frac{64}{125}\right)^{2/3}$

Now, $\left(\frac{64}{125}\right)^{1/3} = \frac{4}{5}$ so $\left(\frac{64}{125}\right)^{2/3} = \left(\frac{4}{5}\right)^2 = \frac{16}{25}$

Hence, $T_X = \frac{16}{25} T_W$

Use isobaric relation for $X \to Y$

For an isobaric process of an ideal gas, $\frac{V}{T} = \text{constant} \Rightarrow T \propto V$

So, $\frac{T_Y}{T_X} = \frac{V_Y}{V_X}$

Given: $V_Y = 250\ \text{cm}^3, \quad V_X = 125\ \text{cm}^3$ Thus, $\frac{T_Y}{T_X} = \frac{250}{125} = 2$ $T_Y = 2T_X$

Therefore, $\Delta T_{XY} = T_Y - T_X = T_X$

Using $T_X = \frac{16}{25}T_W$, $\Delta T_{XY} = \frac{16}{25}T_W$

Heat absorbed in an isobaric process

For a monatomic ideal gas, $C_P = \frac{5R}{2}$

Hence heat absorbed along $X \to Y$ is $Q_{XY} = n C_P (T_Y - T_X)$ $Q_{XY} = n \cdot \frac{5R}{2} \cdot \frac{16}{25}T_W$

Rearranging, $Q_{XY} = \frac{5}{2} \cdot \frac{16}{25} \cdot nRT_W$ $Q_{XY} = \frac{8}{5} \, nRT_W$

Given, $nRT_W = 1\ \text{J}$

Therefore, $Q_{XY} = \frac{8}{5} = 1.6\ \text{J}$

Final answer

The amount of heat absorbed along $X \to Y$ is $\boxed{1.6\ \text{J}}$