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JEE Advanced 2025Paper 2
Exam
JEE Advanced
Year
2025
Paper
Paper 2
Subject
Physics
Chapter
Thermodynamics

JEE Advanced 2025 · Paper 2 · Physics · Thermodynamics

An ideal monatomic gas of n moles is taken through a cycle W X Y Z...

An ideal monatomic gas of n moles is taken through a cycle WXYZWW X Y Z W consisting of consecutive adiabatic and isobaric quasi-static processes, as shown in the schematic V−TV-T diagram. The volume of the gas at W,XW, X and YY points are, 64 cm3,125 cm364 \mathrm{~cm}^3, 125 \mathrm{~cm}^3 and 250 cm3250 \mathrm{~cm}^3, respectively. If the absolute temperature of the gas TWT_W at the point WW is such that nRTW=1 Jn R T_W=1 \mathrm{~J} ( R is the universal gas constant), then the amount of heat absorbed (in J ) by the gas along the path XYX Y is \qquad An ideal monatomic gas of n moles is taken through a cycle W X Y Z W consisting of consecutive adiabatic and isobaric

Correct answer

1.6

Step-by-step solution

nRTW=PWVW=1JPW=164×106Pa\begin{aligned} & \mathrm{nRT}_{\mathrm{W}}=\mathrm{P}_{\mathrm{W}} \mathrm{ V}_{\mathrm{W}}=1 \mathrm{ J} \\ & \mathrm{P}_{\mathrm{W}}=\frac{1}{64} \times 10^6 \mathrm{ Pa} \end{aligned} For WX process PXVXY=PWVWy⇒PX=PW(VWVX)y\begin{aligned} & P_X V_X^Y=P_W V_W^y \\ & \Rightarrow P_X=P_W\left(\frac{V_W}{V_X}\right)^y \end{aligned} amount of heat absorbed in XY process Q=nCPΔT=n×52R×[TY−TX][ For monoatomic gas CP=5R2]Q=52[nRTY−nRTTX]=52[PYVY−PXVX]=52PX[VY−VX][∵PX=PY; Isobaric process ]=52×PW×[VWVX]y[VY−VX]\begin{aligned} \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[\text{ For monoatomic gas } \mathrm{C}_{\mathrm{P}}=\frac{5 \mathrm{R}}{2}\right] \\ \mathrm{Q} & =\frac{5}{2}\left[\mathrm{nRT}_{\mathrm{Y}}-\mathrm{nRT} \mathrm{T}_{\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}} ; \text{ Isobaric process }\right] \\ & =\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]\end{aligned} Putting values : Q=1.6Q=1.6 Joule
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