JEE Advanced 2025 · Paper 2 · Physics Q11

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PhysicsNumerical+4Question 27 of 48

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 $nRT_W = 1$ J ( $R$ is the universal gas constant), then the amount of heat absorbed (in J) by the gas along the path $XY$ is _____.

Schematic V-T diagram (V vertical, T horizontal) of the cycle W→X→Y→Z→W. W is at low V and moderate T; the curved path W→X goes up and to the left (adiabatic, V increases, T decreases). X→Y is a straight segment lying on a line through the origin (isobaric) up to Y at the highest V. Y→Z is a curve down and to the right (adiabatic) to Z at the highest T, and Z→W is a straight segment on a line through the origin (isobaric) back to W. Dotted lines from the origin pass through X and W.

Your answer

Answer and worked solution

Official answer: 1.6

Identify the processes. In the $V$ - $T$ diagram, $XY$ lies on a line through the origin, so it is isobaric. $WX$ is adiabatic.

Temperature at X. For a monatomic gas $\gamma-1=\frac23$ , so
$T_X=T_W\left(\frac{64}{125}\right)^{2/3}=\frac{16}{25}T_W.$

Temperature at Y. Isobaric doubling of volume: $T_Y=2T_X$ .

Heat absorbed along XY.
$Q=\frac52nR(T_Y-T_X)=\frac52\cdot\frac{16}{25}\,nRT_W=1.6\ \text{J}.$

Topics:Thermodynamic processes Specific heats of gases

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JEE Advanced 2025 · Paper 2, Physics question 11