$\text{An ideal monatomic gas of}\mathrm{ n }\text{moles is taken through a cycle}\mathrm{ W}XYZW\text{ consisting of consecutive adiabatic and}$

$\text{isobaric quasi-static processes, as shown in the schematic}\mathrm{ V}-T\text{diagram.}\text{The volume of the gas at}W,X\text{ and}\mathrm{ Y }\text{points}$

$\text{are,}64{\mathrm{cm}}^3,125{\mathrm{cm}}^3\text{and}250{\mathrm{cm}}^3\text{respectively.}\text{If the absolute temperature of the gas at the point}\mathrm{ W}\text{ is such that}\mathrm{ n}$

$\mathrm{ R}{T}_W=1J\left(R\text{is the universal gas constant}\right),\text{then amount of heat absorbed (in J) by the gas along the path}$

$XY\text{is}\text{\_\_\_\_\_\_}$

The correct answer is 1.6.

Step 1: Understand the processes shown in the V-T diagram
In the given volume-temperature (V-T) diagram:
- The processes $WX$ and $YZ$ are adiabatic processes.
- The processes $XY$ and $ZW$ are represented by straight lines pointing toward the origin, which means volume is directly proportional to temperature ($V\propto T$). Thus, these are isobaric (constant pressure) processes.

We are given the following values:
- Volume at point $W$, $V_W=64{\mathrm{cm}}^3$
- Volume at point $X$, $V_X=125{\mathrm{cm}}^3$
- Volume at point $Y$, $V_Y=250{\mathrm{cm}}^3$
- Universal gas constant relation at $W$, $nR{T}_W=1J$

Step 2: Analyze the adiabatic process W-X
For a quasi-static adiabatic process of an ideal gas, the relation between temperature and volume is given by:

$T{V}^{\gamma -1}=\text{constant}$

For an ideal monatomic gas, the ratio of specific heats is:

$\gamma =\frac{5}{3}\Rightarrow \gamma -1=\frac{2}{3}$

Applying this relation between points $W$ and $X$:

$T_W{V}_W^{2/3}=T_X{V}_X^{2/3}$

Solving for $T_X$:

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

Substitute the given volumes:

$T_X=T_W{\left(\frac{64}{125}\right)}^{2/3}=T_W{\left[{\left(\frac{4}{5}\right)}^3\right]}^{2/3}=T_W{\left(\frac{4}{5}\right)}^2=\frac{16}{25}$

Multiplying by $nR$:

$nR{T}_X=\frac{16}{25}nR{T}_W=\frac{16}{25}\left(1\text{J}\right)=0.64\text{J}$

Step 3: Analyze the isobaric process X-Y
Since the process $XY$ is isobaric, we have:

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

Substitute the given volumes:

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

Multiplying by $nR$:

$nR{T}_Y=2nR{T}_X=2\left(0.64\text{J}\right)=1.28\text{J}$

Step 4: Calculate the heat absorbed along path X-Y
The heat absorbed at constant pressure ($Q_{XY}$) is given by:

$Q_{XY}=n{C}_p(T_Y-T_X)$

For a monatomic gas, the molar heat capacity at constant pressure is $C_p=\frac{5}{2}R$. Therefore:

$Q_{XY}=\frac{5}{2}(nR{T}_Y-nR{T}_X)$

Substituting the values of $nR{T}_Y$ and $nR{T}_X$:

$Q_{XY}=\frac{5}{2}(1.28\text{J}-0.64\text{J})=\frac{5}{2}\left(0.64\text{J}\right)=1.6\text{J}$