An ideal gas of density enters a chimney of height at the rate of from its lower end, and escapes through the upper end as shown in the figure. [2022]

Question

An ideal gas of density $ρ = 0.2 {kg m}^{- 3}$ enters a chimney of height $h$ at the rate of $α = 0.8 {kg s}^{- 1}$ 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 $g = 10 {m s}^{- 2}$ and the ratio of specific heats of the gas $γ = 2 .$ Ignore atmospheric pressure.

Which of the following statement(s) is(are) correct [2022]

Which of the following statement(s) is(are) correct? [2022]

Smart Achievers · Accepted Answer

(2)

$As expansion of gas is adiabatic$

$so, P T^{\frac{γ}{1 - γ}} = constant$

$\Rightarrow 600 \times 300^{- 2} = P_{2} \times 150^{- 2}$

$\Rightarrow P_{2} = \frac{600 \times 150^{2}}{300^{2}}$

$\Rightarrow P_{2} = 150 Pa$

$So (1) is incorrect.$

$As \frac{d m}{d t} = ρ A v$

$At 1.$

$0.8 = 0.2 \times 0.1 \times v_{1}$

$v_{1} = 40 m/s$

$By ideal gas equation$

$P V = n R T$

$\Rightarrow P V = \frac{m}{M_{0}} R T \Rightarrow P M_{0} = ρ R T \Rightarrow \frac{ρ}{P T} = constant$

$\Rightarrow \frac{P_{1}}{ρ_{1} T_{1}} = \frac{P_{2}}{ρ_{2} T_{2}} \Rightarrow \frac{600}{0.2 \times 300} = \frac{150}{ρ_{2} \times 150}$

$\Rightarrow ρ = 0.1 {kg/m}^{3}$

$So option (4) is incorrect.$

$Again, \frac{d m}{d t} = ρ A v$

$At 2.$

$0.8 = 0.1 \times 0.4 \times v_{2}$

$\Rightarrow v_{2} = 20 m/s$

$So option (2) is correct.$

$Here, T \neq constant. So Bernoulli's theorem needs to be modified.$

$So, we need to add internal energy per unit volume factor.$

$Therefore, factor of \frac{n C_{v} T}{V} is added.$

$As \frac{n C_{v} T}{V} = \frac{n (\frac{f}{2} R) T}{V} = \frac{n R T}{V} \times \frac{2}{2} = P$

$So, Bernoulli's theorem becomes$ $2 P + \frac{1}{2} ρ v^{2} + ρ g h = constant$

$\Rightarrow 2 P_{1} + \frac{1}{2} ρ_{1} v_{1}^{2} + ρ_{1} g \times 0 = 2 P_{2} + \frac{1}{2} ρ_{2} v_{2}^{2} + ρ_{2} g h$

$\Rightarrow 2 \times 600 + \frac{1}{2} \times 0.2 \times 40^{2} = 2 \times 150 + \frac{1}{2} \times 0.1 \times 20^{2} + 0.1 \times 10 \times h$

$\Rightarrow h = 1040 m$

$So option (3) is incorrect.$

Solution

1 The pressure of the gas at the upper end of the chimney is $300 Pa$ .

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

3 The height of the chimney is $590 m$ .

4 The density of the gas at the upper end is $0.05 {kg m}^{- 3}$ .

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Solution

1 The pressure of the gas at the upper end of the chimney is 300 Pa.

2 The velocity of the gas at the lower end of the chimney is 40 m s-1 and at the upper end is 20 m s-1.

3 The height of the chimney is 590 m.

4 The density of the gas at the upper end is 0.05 kg m-3.

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1 The pressure of the gas at the upper end of the chimney is $300 Pa$ .

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

3 The height of the chimney is $590 m$ .

4 The density of the gas at the upper end is $0.05 {kg m}^{- 3}$ .