An ideal gas of density ρ = 0 . 2kg m ^ - 3 enters a chimney of...

Here the gas undergoes adiabatic expansion,

${P}^{1 - \gamma}{T}^{\gamma} =$constant

$⇒ \dfrac{{P}_{2}}{{P}_{1}} = \left(\dfrac{{T}_{1}}{{T}_{2}}\right)\dfrac{\gamma}{1 - \gamma}$

$⇒ {P}_{2} = \left(\dfrac{300}{150}\right)\dfrac{2}{1 - 2} \times 600$

$⇒ {P}_{2} = \dfrac{600}{4} = 150Pa$

Thus, pressure of the gas at the upper end of the chimney is ${P}_{2} = 150Pa$

From ideal gas equation, $ρ = \dfrac{PM}{RT} ⇒ ρ ∝ \dfrac{P}{T}$

Then, $\dfrac{{ρ}_{1}}{{ρ}_{2}} = \left(\dfrac{{P}_{1}}{{P}_{2}}\right)\left(\dfrac{{T}_{1}}{{T}_{2}}\right) = \left(\dfrac{150}{600}\right)\left(\dfrac{300}{150}\right) = \dfrac{1}{2}$

$⇒ {ρ}_{2} = \left(0 . 2\right)\dfrac{1}{2} = 0 . 1kg{m}^{- 3}$

Rate of mass flow $\alpha = 0 . 8kg{s}^{- 1}$

$⇒$ Volume flow rate at top $= \dfrac{0 . 8}{0 . 1} = 8{m}^{3}{s}^{- 1}$

Velocity of gas at lower end ${v}_{1} = \dfrac{{V}_{1}}{{A}_{1}} = \dfrac{0 . 8}{0 . 2 \times 0 . 1} = 40m{s}^{- 1}$

Velocity of gas at upper end ${v}_{2} = \dfrac{0 . 8 \times 2}{0 . 2 \times 0 . 4} = 20m{s}^{- 1}$

Work done ${W}_{ongas} = ∆ K + ∆ {U}_{g} + \text{internal energy}$

${P}_{1}{A}_{1} ∆ {x}_{1} - {P}_{2}{A}_{2} ∆ {x}_{2} = \dfrac{1}{2} ∆ m{{v}_{1}}^{2} - \dfrac{1}{2} ∆ m{{v}_{2}}^{2} + mg ∆ h + \dfrac{f}{2}\left({P}_{2} ∆ {V}_{2} - {P}_{1} ∆ {V}_{1}\right)$

Solving we get $h = 360m$