An ideal gas of density =0.2 kg m ^ -3 enters a chimney of height… - Vidyadip

MCQ

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 $300 Pa$.
✓
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

Correct option: B.

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

b
$\frac{ dm }{ dt }=\rho_1 A _1 v _1=0.8 kg / sA$

$v _1=\frac{0.8}{0.2 \times 0.1}=40 m / s$

$g=10 m / s ^2$

$\gamma=2$

Gas undergoes adiabatic expansion,

$p ^{1 \gamma} T ^\gamma=\text { Constant }$

$\frac{ P _2}{ P _1}=\left(\frac{ T _1}{ T _2}\right)^{\frac{ T }{1-\gamma}}$

$P _2=\left(\frac{300}{150}\right)^{\frac{2}{-1}} \times 600$

$P _2=\frac{600}{4}=150 Pa$

Now $\rho=\frac{ PM }{ RT } \Rightarrow \rho \propto \frac{ P }{ T }$

$\frac{\rho_1}{\rho_2}=\left(\frac{ P _1}{ P _2}\right)\left(\frac{ T _1}{ T _2}\right)=\left(\frac{150}{600}\right)\left(\frac{300}{150}\right)=\frac{1}{2}$

$\rho_2=\frac{\rho_1}{2}=0.1 kg / m ^3$

$\text { Now } \rho_2 A _2 V _2=0.8 \Rightarrow v _2=\frac{0.8}{0.1 \times 0.4}=20 m / s$

$\text { Now } W _{\text {on gas }}=\Delta K +\Delta U +\text { (Internal energy) }$

$P _1 A _1 \Delta x _1- P _2 A _2 \Delta x _2=\frac{1}{2} \Delta mV_{2 } ^ { 2 } -\frac{1}{2} \Delta mV V _1^2+\Delta mgh +\frac{ f }{2}\left( P _2 \Delta V _2- P _1 \Delta V _1\right)$

$\Rightarrow 2 P _1 \frac{\Delta V _1}{\Delta m }-2 P _2 \frac{\Delta V _2}{\Delta m }=\frac{ V _2^2- V _1^2}{2}+ gh$

$\Rightarrow \frac{2 \times 600}{0.2}-\frac{2 \times 150}{0.1}=\frac{20^2-40^2}{2}+10 h$

$h =360 m$

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