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JEE Advanced Physics Thermodynamics 2022 JEE Advanced 2022 (Paper 1)

JEE Advanced Physics Question (2022) — Solution

Question

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   List-II (I) 10 - 3   kg of water 100   ° C is converted to steam at the same temperature, at a pressure of 10 5 Pa . The volume of the system changes from 10 - 6 m 3 to 10 - 3 m 3 in the process. Latent heat of water  = 2250   kJ   kg - 1 (P) 2   kJ (II) 0 . 2  moles of a rigid diatomic ideal gas with volume V  at temperature 500   K undergoes an isobaric expansion to volume 3   V . Assume  8 . 0   J   mol – 1   K – 1 (Q) 7   kJ (III) One mole of a monoatomic ideal gas is compressed adiabatically from volume V = 1 3 m 3 and pressure 2   kPa to volume 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 Which one of the following options is correct?

Options

  1. A. I → T , II → R , III → S , IV → Q
  2. B. I → S , II → P , III → T , IV → P
  3. C. I → P , II → R , III → T , IV → Q
  4. D. I → Q , II → R , III → S , IV → T

Answer

C. I → P , II → R , III → T , IV → Q

Step-by-step solution

(I) According to the first law of thermodynamics, ∆ Q = ∆ U + W ⇒ ∆ U = ∆ Q - W ⇒ U = M L - P Δ V = 10 - 3 × 2250 - 10 2 kP × 10 - 3 - 10 - 6 m 3 = 2 . 25   kJ - 0 . 1   kJ = 2 . 15   kJ Therefore, I-P (II) For isobaric process, V 1 V 2 = T 1 T 2 ⇒ T 2 = 3 × 500 = 1500   K Now the change in the internal energy will be, ∆ U = n C V ∆ T = 0 . 2 × 5 2 × 8 × 1000 = 4   kJ Therefore,  II-R (III)  For adiabatic expansion of monoatomic gas  γ = 5 3 P 1 V 1 γ = P 2 V 2 γ ⇒ 2 kPa × V 0 5 3 = P 2 × V 0 8 5 3 ⇒ P 2 = 64   kPa Now the change in the internal energy will be, Δ U = n C v Δ T = 3 n R Δ T 2 = 3 2 P 2 V 2 - P 1 V 1 = 3 2 × 64 × 1 3 × 8 - 2 × 1 3 = 3 2 × 8 3 - 2 3 = 3   kJ Therefore,  III-T (IV) For diatomic gas whose molecules vibrate, C V = 6 2 R  and  C P = 8 2 R ⇒ Δ U = n C V Δ T = 3 n R Δ T Δ Q = n C p Δ T = 4 n R Δ T Δ U Δ Q = 3 4 Δ U = 3 4 × 9 = 6 . 75 ≃ 7   kJ Therefore,  IV-Q

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