Question

An ideal gas of density $\rho=0.2 \,kg \, m ^{-3}$ enters a chimney of height $h$ at the rate of $\alpha=0.8 \,kg \,s ^{-1}$ 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 $g=10 \,ms ^{-2}$ 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 $300\, Pa$.
• B. The velocity of the gas at the lower end of the chimney is $40 \,ms ^{-1}$ and at the upper end is $20\, ms ^{-1}$.
• C. The height of the chimney is $590 \,m$.
• D. The density of the gas at the upper end is $0.05\, kg\, m ^{-3}$.

Answer 1

Answer: (B)

$ \frac{ dm }{ dt }=\rho_1 A _1 v _1=0.8 \,kg / s A $
$ v _1=\frac{0.8}{0.2 \times 0.1}=40 \,m / s $
$ g =10 \,m / 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}=150 \,Pa$
Now $\rho=\frac{ PM }{ RT } \Rightarrow \rho \propto \frac{ P }{ T }$
$\frac{\rho_1}{\rho_2}=\left(\frac{P_1}{P_2}\right)\left(\frac{T_1}{T_2}\right)=\left(\frac{150}{600}\right)\left(\frac{300}{150}\right)=\frac{1}{2}$
$\rho_2=\frac{\rho_1}{2}=0.1 \,kg / m ^3$
Now $\rho_2 A _2 v _2=0.8 \Rightarrow v _2=\frac{0.8}{0.1 \times 0.4}=20\, m / s$
Now $W _{\text {on gas }}=\Delta K +\Delta U +($ Internal energy $)$
$P _1 A _1 \Delta x _1- P _2 A _2 \Delta x _2=\frac{1}{2} \Delta mV _2^2-\frac{1}{2} \Delta mV V _1^2+\Delta mgh +\frac{ f }{2}\left( P _2 \Delta V _2- P _1 \Delta V _1\right)$
$\Rightarrow 2 P _1 \frac{\Delta V _1}{\Delta m }-2 P _2 \frac{\Delta V _2}{\Delta m }=\frac{ V _2^2- V _1^2}{2}+ gh$
$\Rightarrow \frac{2 \times 600}{0.2}-\frac{2 \times 150}{0.1}=\frac{20^2-40^2}{2}+10 h$
$h =360\, m$