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← JEE Advanced 2022 · Paper 2
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JEE Advanced 2022 · Paper 2, Physics question 14

In the given PP-VV diagram, a monoatomic gas (γ=53)\left(\gamma = \frac53\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\left(\frac13\right)^{0.6} \simeq 0.5, ln⁡2≃0.7\ln 2 \simeq 0.7].

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

P-V diagram: pressure P (kPa) on the vertical axis, volume V (m^3) on the horizontal axis. State A is at V = 0.80 m^3, P = 100 kPa. An adiabatic compression curve (arrow) runs from A up and to the left to state B at P = 300 kPa (dashed horizontal line at 300 kPa; B's volume is not marked). From B an isothermal expansion curve (arrow) runs up and to the right to state C, which lies directly above A on the vertical line V = 0.80 m^3. Dashed guide lines mark P = 100 and P = 300 kPa.
Choose one or more options
Answer and worked solution

Official answer: Dropped: full marks to everyone

This question was dropped (marks to all) in the official key. Below is the standard working.

States.

  • AA: (0.8 m3, 100(0.8\ \mathrm{m^3},\ 100 kPa).
  • Adiabatic to 300 kPa: VB=0.8(13)1/γ=0.8(0.5)=0.4 m3V_B=0.8\left(\frac13\right)^{1/\gamma}=0.8(0.5)=0.4\ \mathrm{m^3}, so PBVB=120P_BV_B=120 kJ.
  • Isothermal to VC=0.8 m3V_C=0.8\ \mathrm{m^3}.

Work in each process.

  • A→BA\to B: ∣W∣=PBVB−PAVAγ−1=120−802/3=60|W|=\dfrac{P_BV_B-P_AV_A}{\gamma-1}=\dfrac{120-80}{2/3}=60 kJ, done on the gas. (C) is true.
  • B→CB\to C: ∣W∣=PBVBln⁡VCVB=120ln⁡2=84|W|=P_BV_B\ln\dfrac{V_C}{V_B}=120\ln2=84 kJ, done by the gas. (B) is true.
  • C→AC\to A: the volume is constant, so W=0W=0. (D) is true.

Option (A). The net work for A→B→CA\to B\to C is 84−60=2484-60=24 kJ. Adding the magnitudes gives 60+84=14460+84=144 kJ. Read literally, (A) is false, but "magnitude of the total work" can be read either way, which is probably why the question was dropped.