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JEE ADVANCED 2025
Paper-2 2025
Question
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,$64 \mathrm{~cm}^3, 125 \mathrm{~cm}^3$ and $250 \mathrm{~cm}^3$ respectively. If the absolute temperature of the gas $T_W$ at the point $W$ is such that $n B J_\alpha=1 J(R$ is the universal gas constant), then the amount of heat absorbed (in $J)$ by the gas along the path $X Y$ is $\_\_\_\_$
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Solution

$\begin{aligned} & Q_{x y}=P_1\left(V_y-V_x\right)+n R\left(T_y-T_x\right) \frac{3}{2} \\ & Q_{x y}=\frac{5}{2} R\left(T_y-T_x\right) n \\ & T_{x y} \cdot(64)^{\frac{2}{3}}=T_x(125)^{\frac{2}{3}} \\ & T_x \cdot 16=T_x \cdot 25 \\ & T_x=\frac{16}{25} T_w \\ & \frac{T_y}{V_y}=\frac{T_x}{V_x} \\ & T_y=T_x\left(\frac{250}{125}\right)=2 T_x \\ & Q_{x y}=\frac{5}{2} n R\left(T_x\right)=\frac{5}{2} n R \cdot \frac{16}{25} T_w=\frac{8}{5} \\ & Q_x=1.6\end{aligned}$
Question Tags
JEE Advance
Physics
Easy
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