Question

An ideal gas of density $\rho =0.2kg{m}^{-3}$ enters a chimney of height $h$ at the rate of $\alpha =0.8kg{s}^{-1}$ from its lower end, and escapes through the upper end as shown in the figure. The cross-sectional area of the lower end is $A_1=0.1m^2$ and the upper end is $A_2=0.4m^2$. The pressure and the temperature of the gas at the lower end are $600Pa$ and $300K$, respectively, while its temperature at the upper end is $150K$. The chimney is heat insulated so that the gas undergoes adiabatic expansion. Take $g=10m{s}^{-2}$ and the ratio of specific heats of the gas $\gamma =2$. Ignore atmospheric pressure.

Which of the following statement(s) is(are) correct?

• A. The pressure of the gas at the upper end of the chimney is $300Pa$.
• B. The velocity of the gas at the lower end of the chimney is $40m{s}^{-1}$ and at the upper end is $20m{s}^{-1}$.
• C. The height of the chimney is $590m$.
• D. The density of the gas at the upper end is $0.05kg{m}^{-3}$.

Answer 1

Answer: (B)

$\frac{dm}{dt}={\rho }_1{A}_1{v}_1=0.8kg/sA$
$v_1=\frac{0.8}{0.2\times 0.1}=40m/s$
$g=10m/s^2$
$\gamma =2$
Gas undergoes adiabatic expansion,
$p^{1-\gamma }{T}^{\gamma }=\text{ Constant }$
$\frac{P_2}{P_1}={\left(\frac{T_1}{T_2}\right)}^{\frac{r}{1-\gamma }}$
$P_2={\left(\frac{300}{150}\right)}^{\frac{2}{-1}}\times 600$
$P_2=\frac{600}{4}=150Pa$
Now $\rho =\frac{PM}{RT}\Rightarrow \rho \propto \frac{P}{T}$
$\frac{{\rho }_1}{{\rho }_2}=\left(\frac{P_1}{P_2}\right)\left(\frac{T_1}{T_2}\right)=\left(\frac{150}{600}\right)\left(\frac{300}{150}\right)=\frac{1}{2}$
${\rho }_2=\frac{{\rho }_1}{2}=0.1kg/m^3$
Now ${\rho }_2{A}_2{v}_2=0.8\Rightarrow v_2=\frac{0.8}{0.1\times 0.4}=20m/s$
Now $W_{\text{on gas }}=\Delta K+\Delta U+($ Internal energy $)$
$P_1{A}_1\Delta x_1-P_2{A}_2\Delta x_2=\frac{1}{2}\Delta m{V}_2^2-\frac{1}{2}\Delta mV{V}_1^2+\Delta mgh+\frac{f}{2}\left(P_2\Delta V_2-P_1\Delta V_1\right)$
$\Rightarrow 2P_1\frac{\Delta V_1}{\Delta m}-2P_2\frac{\Delta V_2}{\Delta m}=\frac{V_2^2-V_1^2}{2}+gh$
$\Rightarrow \frac{2\times 600}{0.2}-\frac{2\times 150}{0.1}=\frac{20^2-40^2}{2}+10h$
$h=360m$