An ideal monatomic gas of n moles is taken through a cycle WXYZW consisting of consecutive adiabatic and isobaric quasi-static processes, as shown in the schematic $V-T$ diagram. The volume of the gas at W, X and Y points are 64 cm$^{3}, 125$ cm$^{3}$ and 250 cm$^{3},$ respectively. If the absolute temperature of the gas $T_{W}$ at the point W is such that nRT$_{W} = 1 J (R$ is the universal gas constant), then the amount of heat absorbed (in J) by the gas along the path XY is ______.

Step 1: Identify the processes and gas properties.

For an ideal monatomic gas, the ratio of specific heats is $\gamma = \frac{5}{3}$, which gives:

$$\gamma - 1 = \frac{2}{3}$$

The processes $W \to X$ and $Y \to Z$ are adiabatic, while $X \to Y$ and $Z \to W$ are isobaric.

Step 2: Analyze the adiabatic process $W \to X$.

Using the relation $T V^{\gamma-1} = \text{constant}$ for an adiabatic process:

$$T_W V_W^{2/3} = T_X V_X^{2/3}$$ $$T_X = T_W \left(\frac{V_W}{V_X}\right)^{2/3}$$

Substitute the volumes $V_W = 64\,\text{cm}^{3}$ and $V_X = 125\,\text{cm}^{3}$:

$$T_X = T_W \left(\frac{64}{125}\right)^{2/3}$$ $$T_X = \frac{16}{25} T_W$$

Step 3: Analyze the isobaric process $X \to Y$.

For an isobaric process, the temperature is directly proportional to the volume:

$$\frac{T_Y}{V_Y} = \frac{T_X}{V_X}$$ $$T_Y = T_X \left(\frac{V_Y}{V_X}\right)$$

Substitute the volumes $V_Y = 250\,\text{cm}^{3}$ and $V_X = 125\,\text{cm}^{3}$:

$$T_Y = T_X \left(\frac{250}{125}\right)$$ $$T_Y = 2 T_X$$ $$T_Y = \frac{32}{25} T_W$$

Step 4: Calculate the heat absorbed along the path $XY$.

The heat absorbed during an isobaric process is given by $Q = n C_p \Delta T$.

$$C_p = \frac{5}{2}R$$ $$Q_{XY} = n \left(\frac{5}{2}R\right) (T_Y - T_X)$$ $$Q_{XY} = \frac{5}{2} nR \left(\frac{32}{25} T_W - \frac{16}{25} T_W\right)$$ $$Q_{XY} = \frac{5}{2} nR \left(\frac{16}{25} T_W\right)$$ $$Q_{XY} = \frac{8}{5} nRT_W$$

Step 5: Substitute the given value of $nRT_W$.

Substitute $nRT_W = 1\,\text{J}$ into the heat equation:

$$Q_{XY} = \frac{8}{5} (1\,\text{J})$$ $$Q_{XY} = 1.6\,\text{J}$$

Answer: $1.6$