JEE Advanced 2025 — Paper 2 — Question 11

$n R T_{w}=P_{w} V_{w}=1 \mathrm{~J}$

$\mathrm{P}_{\mathrm{W}}=\frac{1}{64} \times 10^{6} \mathrm{~Pa}$

For WX process $\mathrm{P}_{\mathrm{X}} \mathrm{V}_{\mathrm{X}}^{\mathrm{y}}=\mathrm{P}_{\mathrm{W}} \mathrm{V}_{\mathrm{W}}^{\mathrm{y}}$

$\Rightarrow P_{X}=P_{W}\left(\frac{V_{W}}{V_{X}}\right)^{y}$

amount of heat absorbed in XY process $\mathrm{Q}=\mathrm{nCP} \Delta \mathrm{T}=\mathrm{n} \times \frac{5}{2} \mathrm{R} \times\left[\mathrm{T}_{\mathrm{Y}}-\mathrm{T}_{\mathrm{X}}\right] \quad\left[\right.$

For monoatomic gas $\left.\mathrm{C}_{\mathrm{P}}=\frac{5 \mathrm{R}}{2}\right]$

$\mathrm{Q}=\frac{5}{2}\left[\mathrm{nRT}_{\mathrm{Y}}-\mathrm{nRT}_{\mathrm{X}}\right]$

$=\frac{5}{2}\left[\mathrm{P}_{\mathrm{Y}} \mathrm{V}_{\mathrm{Y}}-\mathrm{P}_{\mathrm{X}} \mathrm{V}_{\mathrm{X}}\right]$

$=\frac{5}{2} \mathrm{P}_{\mathrm{X}}\left[\mathrm{V}_{\mathrm{Y}}-\mathrm{V}_{\mathrm{X}}\right] \quad\left[\because \quad \mathrm{P}_{\mathrm{X}}=\mathrm{P}_{\mathrm{Y}} ;\right.$ Isobaric process $]$

$=\frac{5}{2} \times \mathrm{P}_{\mathrm{W}} \times\left[\frac{\mathrm{V}_{\mathrm{W}}}{\mathrm{V}_{\mathrm{X}}}\right]^{\mathrm{Y}}\left[\mathrm{V}_{\mathrm{Y}}-\mathrm{V}_{\mathrm{X}}\right]$\ Putting values : Q = 1.6 Joule