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

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 .