Question

An ideal gas of density $\rho=02 \,kg \, m ^{-3}$ enters a chimney of height $h$ at the rate of $\alpha=08 \,kg \,s ^{-1}$ from its lower end, and escapes through the upper end as shown in the figure The cross-sectional area of the lower end is $A_1=01 \, m ^2$ and the upper end is $A_2=04 \, m ^2$ 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

The Correct Option is B

\(\frac{dm}{dt} = p_1 A_1 v_1 = 0.8 \, \text{kg/s} \, A\)
\(v_1 = \frac{0.8}{0.2 \times 0.1} = 40 \, \text{m/s}\)

g = 10 m/s² 
\(\gamma = 2\)

Gas undergoes adiabatic expansion,
\(p_1 - \gamma T^{\gamma} = \text{Constant}\)
\(\frac{P_2}{P_1}=(\frac{T_1}{T_2})^{\frac{r}{1-\gamma}}\)

\(P_2=\frac{600}{4}=150P_a\)

Now \(\rho∝\frac{P}{T}\)

\(\frac{\rho_1}{\rho_2}=(\frac{P_1}{P_2})(\frac{T_1}{T_2})\)

\((\frac{150}{600})(\frac{300}{150})=\frac{1}{2}\)

Now
\(P_1 A_1 \Delta x_1 - P_2 A_2 \Delta x_2 = 2 mV_1 - mV_2 + mgh_2 - mgh_1 + \frac{f}{2}(P_2 V_2 - P_1 V_1)\)

Simplifying we get.  \(\frac{V_2}{V_1} - \frac{V_1}{V_2} = \frac{2P}{gh} \frac{m}{m}\)
\(⇒\frac{2\times600}{0.2}-\frac{2\times150}{0.1}\)

\(=\frac{20^2-40^2}{2}+10h\)
\(h = 360 m\)