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Physics · Thermodynamics

JEE Advanced 2022 — Paper 1 — Question 33

List I describes thermodynamic processes in four different systems. List II gives the magnitudes (either exactly or as a close approximation) of possible changes in the internal energy of the system due to the process.

List −I-\mathbf{I}List -II
(I)A mass of 10−3 kg10^{-3}\,\text{kg} of water at 100∘C100^\circ\text{C} is converted into steam at the same temperature under a pressure of 105 Pa10^{5}\,\text{Pa}. During the process, the volume of the system changes from 10−6 m310^{-6}\,\text{m}^3 to 10−3 m310^{-3}\,\text{m}^3. The latent heat of vaporisation of water is L=2250 kJ kg−1L = 2250\,\text{kJ}\,\text{kg}^{-1}(P)2 kJ
(II)A sample of 0.20.2 moles of a diatomic ideal gas contained in a rigid vessel of volume VV is at a temperature of 500 K500\,\text{K}. The gas undergoes an isobaric expansion until its volume becomes 3V3V. Take the universal gas constant as R=8.0 J mol−1K−1.R = 8.0\,\text{J mol}^{-1}\text{K}^{-1}.(Q)7 kJ
(III)One mole of a monatomic ideal gas is compressed adiabatically from an initial state of volume V=13 m3 V=\frac{1}{3}\,\text{m}^3 and pressure P=2 kPaP=2\,\text{kPa} to a final volume Vf=V8V_f=\frac{V}{8}(R)4 kJ
(IV)Three moles of a diatomic ideal gas whose molecules can vibrate, is given 9 kJ of heat and undergoes isobaric expansion.(S)5 kJ
(T)3 kJ
  1. Option A:

    I →\rightarrow T, II →\rightarrow R, III →\rightarrow S, IV →Q\rightarrow \mathrm{Q}

  2. Option B:

    I→S,II→P,III→T,IV→P\quad \mathrm{I} \rightarrow \mathrm{S}, \mathrm{II} \rightarrow \mathrm{P}, \mathrm{III} \rightarrow \mathrm{T}, \mathrm{IV} \rightarrow \mathrm{P}

  3. Option C:

    I →\rightarrow P, II →\rightarrow R, III →\rightarrow T, IV →Q\rightarrow \mathrm{Q}

    Correct
  4. Option D:

    I→Q,II→R\quad \mathrm{I} \rightarrow \mathrm{Q}, \mathrm{II} \rightarrow \mathrm{R}, III →S,IV→T\rightarrow \mathrm{S}, \mathrm{IV} \rightarrow \mathrm{T}

Answer: C

Step-by-step solution

(I) Q=mL=10−3×2250=2.25 kJ≈2 kJQ = mL = 10^{-3}\times 2250 = 2.25\,\text{kJ} \approx 2\,\text{kJ}   ⇒  (I)→(P)\;\Rightarrow\; (I)\to(P)

(II) Isobaric ⇒Tf=3Ti⇒ΔT=1000 K\Rightarrow T_f = 3T_i \Rightarrow \Delta T = 1000\,\text{K}

ΔU=nCvΔT=0.2×52×8×1000=4 kJ⇒(II)→(R)\Delta U = nC_v\Delta T = 0.2\times\frac{5}{2}\times 8 \times 1000 = 4\,\text{kJ} \Rightarrow (II)\to(R)

(III) γ=53,  Vf=V8\gamma=\frac{5}{3},\; V_f=\frac{V}{8}

Tf=Ti(ViVf)γ−1=4TiT_f = T_i\left(\frac{V_i}{V_f}\right)^{\gamma-1} = 4T_i

Ti=PVR≈83 K⇒ΔT≈250 KT_i=\dfrac{PV}{R}\approx83\,\text{K} \Rightarrow \Delta T\approx250\,\text{K}

ΔU=32RΔT≈3 kJ⇒(III)→(T)\Delta U = \frac{3}{2}R\Delta T \approx 3\,\text{kJ} \Rightarrow (III)\to(T)

(IV) Cp=92R,  Q=9 kJC_p=\frac{9}{2}R,\; Q=9\,\text{kJ}

ΔT=QnCp≈83 K\Delta T = \frac{Q}{nC_p}\approx83\,\text{K}

W=nRΔT≈2 kJ⇒(IV)→(Q)W = nR\Delta T \approx 2\,\text{kJ} \Rightarrow (IV)\to(Q)

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2022
Paper
Paper 1
Subject
Physics
Chapter
Thermodynamics
Topic
Different Thermodynamic Processes

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List I describes thermodynamic processes in four different systems.… | JEE Advanced 2022 PYQ with Solution · DhiX AI