An ideal gas of density rho_1=0.2 kg m^{-3} | vedclass

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(A, B, C, D) Mass flow rate: $\frac{dm}{dt} = ho_1 A_1 v_1 = 0.8 \ kg/s$.Velocity at lower end: $v_1 = \frac{0.8}{0.2 	imes 0.1} = 40 \ m/s$.For ad

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The velocity of the gas at the lower end of the chimney is $40 \ ms^{-1}$ and at the upper end is $20 \ ms^{-1}$.

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(A, B, C, D) Mass flow rate: $\frac{dm}{dt} = ho_1 A_1 v_1 = 0.8 \ kg/s$.Velocity at lower end: $v_1 = \frac{0.8}{0.2 	imes 0.1} = 40 \ m/s$.For adiabatic expansion: $P^{1-\gamma} T^{\gamma} = 	ext{constant}$,so $\frac{P_2}{P_1} = \left(\frac{T_1}{T_2}ight)^{\frac{\gamma}{\gamma-1}}$.Given $\gamma=2$,$\frac{P_2}{P_1} = \left(\frac{300}{150}ight)^2 = 4 \implies P_2 = 600 	imes \frac{1}{4} = 150 \ Pa$.Using ideal gas law $ho = \frac{PM}{RT}$,$\frac{ho_1}{ho_2} = \left(\frac{P_1}{P_2}ight)\left(\frac{T_2}{T_1}ight) = \left(\frac{600}{150}ight)\left(\frac{150}{300}ight) = 2 \implies ho_2 = \frac{0.2}{2} = 0.1 \ kg/m^3$.Velocity at upper end: $v_2 = \frac{0.8}{ho_2 A_2} = \frac{0.8}{0.1 	imes 0.4} = 20 \ m/s$.Applying Bernoulli's principle for compressible flow (energy conservation): $\frac{\gamma}{\gamma-1} \frac{P_1}{ho_1} + \frac{1}{2}v_1^2 = \frac{\gamma}{\gamma-1} \frac{P_2}{ho_2} + \frac{1}{2}v_2^2 + gh$.$\frac{2}{1} \frac{600}{0.2} + \frac{1}{2}(40)^2 = \frac{2}{1} \frac{150}{0.1} + \frac{1}{2}(20)^2 + 10h$.$6000 + 800 = 3000 + 200 + 10h \implies 6800 = 3200 + 10h \implies 10h = 3600 \implies h = 360 \ m$.

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