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

JEE Advanced Physics Question (2022) — Solution

Question

An ideal gas of density ρ = 0 . 2   kg   m - 3 enters a chimney of height h at the rate of α = 0 . 8   kg   s - 1 from its lower end, and escapes through the upper end as shown in the figure. The cross-sectional area of the lower end is A 1 = 0 . 1   m 2 and the upper end is A 2 = 0 . 4   m 2 . The pressure and the temperature of the gas at the lower end are 600   Pa and 300   K , respectively, while its temperature at the upper end is 150   K . The chimney is heat insulated so that the gas undergoes adiabatic expansion. Take g = 10   m   s - 2 and the ratio of specific heats of the gas γ = 2 . Ignore atmospheric pressure. Which of the following statement(s) is(are) correct?

Options

  1. A. The pressure of the gas at the upper end of the chimney is 300   Pa .
  2. B. The velocity of the gas at the lower end of the chimney is 40   m   s - 1 and at the upper end is 20   m   s - 1 .
  3. C. The height of the chimney is 590   m .
  4. D. The density of the gas at the upper end is 0 . 05   kg   m - 3 .

Answer

B. The velocity of the gas at the lower end of the chimney is 40   m   s - 1 and at the upper end is 20   m   s - 1 .

Step-by-step solution

Here the gas undergoes adiabatic expansion,  P 1 − γ T γ = constant ⇒ P 2 P 1 = T 1 T 2 γ 1 − γ ⇒ P 2 = 300 150 2 1 − 2 × 600 ⇒ P 2 = 600 4 = 150   Pa Thus, pressure of the gas at the upper end of the chimney is  P 2 = 150   Pa From ideal gas equation,  ρ = PM RT ⇒ ρ ∝ P T Then,  ρ 1 ρ 2 = P 1 P 2 T 1 T 2 = 150 600 300 150 = 1 2 ⇒ ρ 2 = 0 . 2 1 2 = 0 . 1   kg   m - 3 Rate of mass flow  α = 0 . 8   kg   s - 1 ⇒ Volume flow rate at top  = 0 . 8 0 . 1 = 8   m 3   s - 1 Velocity of gas at lower end  v 1 = V 1 A 1 = 0 . 8 0 . 2 × 0 . 1 = 40   m   s - 1 Velocity of gas at upper end  v 2 = 0 . 8 × 2 0 . 2 × 0 . 4 = 20   m   s - 1 Work done   W o n   g a s = ∆ K + ∆ U g + internal energy P 1 A 1 ∆ x 1 - P 2 A 2 ∆ x 2 = 1 2 ∆ m v 1 2 - 1 2 ∆ m v 2 2 + m g ∆ h + f 2 P 2 ∆ V 2 - P 1 ∆ V 1 Solving we get  h = 360   m

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