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QJEEAdv22Ph9ade: JEE Advanced 2022 Physics - Class 11 JEEAdv22Ph9ade Solution: In the given diagram, a monoatomic gas is first compressed…

In the given P−VP-V diagram, a monoatomic gas (γ=53)\left(\gamma=\frac{5}{3}\right) is first compressed adiabatically from state AA to state BB . Then it expands isothermally from state BB to state CC . [Given: (13)0.6≃0.5,ln⁡2≃0.7\left(\frac{1}{3}\right)^{0.6} \simeq 0.5, \ln 2 \simeq 0.7 ].

JEE Advanced 2022 Paper 2 Online Physics - Heat and Thermodynamics Question 18 English

Which of the following statement(s) is(are) correct?

Multiple Choice Options

  • A The magnitude of the total work done in the process A→B→CA \rightarrow B \rightarrow C is 144 kJ144 \mathrm{~kJ} .
  • B The magnitude of the work done in the process B→CB \rightarrow C is 84 kJ84 \mathrm{~kJ} .
  • C The magnitude of the work done in the process A→BA \rightarrow B is 60 kJ60 \mathrm{~kJ} .
  • D The magnitude of the work done in the process C→AC \rightarrow A is zero.

Step-by-Step Verified Solution

For adiabatic process (A→B)(\mathrm{A} \rightarrow \mathrm{B})

amp;PAVAγ=PBVBγamp;105×(0.8)53=3×105( VB)53amp;⇒VB=0.8×(13)0.6=0.4\begin{aligned}& \mathrm{P}_{\mathrm{A}} \mathrm{V}_{\mathrm{A}}^\gamma=\mathrm{P}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}^\gamma \\\\& 10^5 \times(0.8)^{\frac{5}{3}}=3 \times 10^5\left(\mathrm{~V}_{\mathrm{B}}\right)^{\frac{5}{3}} \\\\& \Rightarrow \mathrm{V}_{\mathrm{B}}=0.8 \times\left(\frac{1}{3}\right)^{0.6}=0.4\end{aligned}

Work done in process A→B\mathrm{A} \rightarrow \mathrm{B}

amp;WAB=PAVA−PBVBγ−1amp;⇒WAB=105×0.8−3×105×0.453−1amp;⇒WAB=−60 kJ=⇒∣WAB∣=60 kJ\begin{aligned}& \mathrm{W}_{\mathrm{AB}}=\frac{\mathrm{P}_{\mathrm{A}} \mathrm{V}_{\mathrm{A}}-\mathrm{P}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}}{\gamma-1} \\\\& \Rightarrow \mathrm{W}_{\mathrm{AB}}=\frac{10^5 \times 0.8-3 \times 10^5 \times 0.4}{\frac{5}{3}-1} \\\\& \Rightarrow \mathrm{W}_{\mathrm{AB}}=-60 \mathrm{~kJ}=\Rightarrow\left|\mathrm{W}_{\mathrm{AB}}\right|=60 \mathrm{~kJ}\end{aligned}

Work done in process B→C\mathrm{B} \rightarrow \mathrm{C} (Isothermal process)

amp;WBC=nRTℓnVCVB=PBVBℓnVCVBamp;⇒WBC=3×105×0.4ℓn0.80.4amp;⇒WBC=84 kJ\begin{aligned}& \mathrm{W}_{\mathrm{BC}}=\mathrm{nRT} \ell \mathrm{n} \frac{\mathrm{V}_{\mathrm{C}}}{\mathrm{V}_{\mathrm{B}}}=\mathrm{P}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}} \ell \mathrm{n} \frac{\mathrm{V}_{\mathrm{C}}}{\mathrm{V}_{\mathrm{B}}} \\\\& \Rightarrow \mathrm{W}_{\mathrm{BC}}=3 \times 10^5 \times 0.4 \ell \mathrm{n} \frac{0.8}{0.4} \\\\& \Rightarrow \mathrm{W}_{\mathrm{BC}}=84 \mathrm{~kJ}\end{aligned}

Work done in process C→A\mathrm{C} \rightarrow \mathrm{A}

WCA=PΔV=0(∵ΔV=0)\mathrm{W}_{\mathrm{CA}}=\mathrm{P} \Delta \mathrm{V}=0 \quad(\because \Delta \mathrm{V}=0)

So total work done in the process A→B→C\mathrm{A} \rightarrow \mathrm{B} \rightarrow \mathrm{C}

amp;WABC=WAB+WBC+WCA=−60+84+0amp; WABC=24 kJ\begin{aligned}& \mathrm{W}_{\mathrm{ABC}}=\mathrm{W}_{\mathrm{AB}}+\mathrm{W}_{\mathrm{BC}}+\mathrm{W}_{\mathrm{CA}}=-60+84+0 \\\\& \mathrm{~W}_{\mathrm{ABC}}=24 \mathrm{~kJ}\end{aligned}

So correct options are (B,C,D)

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