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 ~m ~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?

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

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

The height of the chimney is 590 ~m .

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

Properties of Matter PYQ — JEE Advanced 2022 · Shift 1 · Q47 with Solution

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Correct answer: B

Given data

Density at lower end: $\rho_1 = 0.2\,\text{kg m}^{-3}$
Mass flow rate: $\dot m = \alpha = 0.8\,\text{kg s}^{-1}$
Area at lower end: $A_1 = 0.1\,\text{m}^2$
Area at upper end: $A_2 = 0.4\,\text{m}^2$
Pressure at lower end: $P_1 = 600\,\text{Pa}$
Temperature at lower end: $T_1 = 300\,\text{K}$
Temperature at upper end: $T_2 = 150\,\text{K}$
$g = 10\,\text{m s}^{-2}$
$\gamma = 2$

Since the chimney is insulated, the flow is adiabatic.

Velocity at lower and upper ends using continuity

For steady flow,

\dot m = \rho A v

Lower end

v_1 = \frac{\dot m}{\rho_1 A_1} = \frac{0.8}{0.2\times 0.1} = \frac{0.8}{0.02} = 40\,\text{m s}^{-1}

So lower-end velocity is $40\,\text{m s}^{-1}$ .

Upper end density

For an ideal gas,

\frac{P}{\rho T}=\text{constant}

and for adiabatic process,

T\rho^{1-\gamma}=\text{constant}

Since $\gamma=2$ ,

T\rho^{-1}=\text{constant} \quad \Rightarrow \quad \frac{T}{\rho}=\text{constant}

Thus,

\frac{T_1}{\rho_1} = \frac{T_2}{\rho_2}

\rho_2 = \rho_1\frac{T_2}{T_1} = 0.2\times \frac{150}{300} = 0.1\,\text{kg m}^{-3}

Then upper-end velocity:

v_2 = \frac{\dot m}{\rho_2 A_2} = \frac{0.8}{0.1\times 0.4} = \frac{0.8}{0.04} = 20\,\text{m s}^{-1}

So statement B is correct.

Pressure at upper end

For adiabatic relation:

T^\gamma P^{1-\gamma}=\text{constant}

For $\gamma=2$ ,

T^2 P^{-1}=\text{constant} \quad \Rightarrow \quad \frac{T^2}{P}=\text{constant}

Hence,

\frac{T_1^2}{P_1} = \frac{T_2^2}{P_2}

P_2 = P_1\left(\frac{T_2}{T_1}\right)^2 = 600\left(\frac{150}{300}\right)^2 = 600\times \frac{1}{4} = 150\,\text{Pa}

So the pressure at the upper end is $150\,\text{Pa}$ , not $300\,\text{Pa}$ .

Therefore, A is false.

Height of the chimney using Bernoulli for adiabatic flow

For steady adiabatic flow,

\frac{v^2}{2} + gz + C_pT = \text{constant}

So between lower end ( $z_1=0$ ) and upper end ( $z_2=h$ ):

\frac{v_1^2}{2} + C_pT_1 = \frac{v_2^2}{2} + gh + C_pT_2

Thus,

gh = \frac{v_1^2-v_2^2}{2} + C_p(T_1-T_2)

Now,

C_p = \frac{\gamma R}{\gamma-1}

We first find $R$ using ideal gas law at lower end:

P_1 = \rho_1 R T_1

R = \frac{P_1}{\rho_1 T_1} = \frac{600}{0.2\times 300} = \frac{600}{60}=10\,\text{J kg}^{-1}\text{K}^{-1}

Hence,

C_p = \frac{2\times 10}{2-1} = 20\,\text{J kg}^{-1}\text{K}^{-1}

Now calculate:

\frac{v_1^2-v_2^2}{2} = \frac{40^2-20^2}{2} = \frac{1600-400}{2} = 600

C_p(T_1-T_2)=20(300-150)=20\times 150=3000

Therefore,

gh = 600+3000=3600

h = \frac{3600}{10}=360\,\text{m}

So statement C is false.

Density at upper end

We already found:

\rho_2 = 0.1\,\text{kg m}^{-3}

So statement D is false.

Final evaluation of options

A: False
B: True
C: False
D: False

Hence, the correct option is:

\boxed{\text{B only}}

Comparison with stored correct answer

Stored correct answer: B

My derived answer: B only

They agree.

More from Properties of Matter

View written solutionFree

Correct answer: B

Given data

Density at lower end: $\rho_1 = 0.2\,\text{kg m}^{-3}$
Mass flow rate: $\dot m = \alpha = 0.8\,\text{kg s}^{-1}$
Area at lower end: $A_1 = 0.1\,\text{m}^2$
Area at upper end: $A_2 = 0.4\,\text{m}^2$
Pressure at lower end: $P_1 = 600\,\text{Pa}$
Temperature at lower end: $T_1 = 300\,\text{K}$
Temperature at upper end: $T_2 = 150\,\text{K}$
$g = 10\,\text{m s}^{-2}$
$\gamma = 2$

Since the chimney is insulated, the flow is adiabatic.

Velocity at lower and upper ends using continuity

For steady flow,

\dot m = \rho A v

Lower end

v_1 = \frac{\dot m}{\rho_1 A_1} = \frac{0.8}{0.2\times 0.1} = \frac{0.8}{0.02} = 40\,\text{m s}^{-1}

So lower-end velocity is $40\,\text{m s}^{-1}$ .

Upper end density

For an ideal gas,

\frac{P}{\rho T}=\text{constant}

and for adiabatic process,

T\rho^{1-\gamma}=\text{constant}

Since $\gamma=2$ ,

T\rho^{-1}=\text{constant} \quad \Rightarrow \quad \frac{T}{\rho}=\text{constant}

Thus,

\frac{T_1}{\rho_1} = \frac{T_2}{\rho_2}

\rho_2 = \rho_1\frac{T_2}{T_1} = 0.2\times \frac{150}{300} = 0.1\,\text{kg m}^{-3}

Then upper-end velocity:

v_2 = \frac{\dot m}{\rho_2 A_2} = \frac{0.8}{0.1\times 0.4} = \frac{0.8}{0.04} = 20\,\text{m s}^{-1}

So statement B is correct.

Pressure at upper end

For adiabatic relation:

T^\gamma P^{1-\gamma}=\text{constant}

For $\gamma=2$ ,

T^2 P^{-1}=\text{constant} \quad \Rightarrow \quad \frac{T^2}{P}=\text{constant}

Hence,

\frac{T_1^2}{P_1} = \frac{T_2^2}{P_2}

P_2 = P_1\left(\frac{T_2}{T_1}\right)^2 = 600\left(\frac{150}{300}\right)^2 = 600\times \frac{1}{4} = 150\,\text{Pa}

So the pressure at the upper end is $150\,\text{Pa}$ , not $300\,\text{Pa}$ .

Therefore, A is false.

Height of the chimney using Bernoulli for adiabatic flow

For steady adiabatic flow,

\frac{v^2}{2} + gz + C_pT = \text{constant}

So between lower end ( $z_1=0$ ) and upper end ( $z_2=h$ ):

\frac{v_1^2}{2} + C_pT_1 = \frac{v_2^2}{2} + gh + C_pT_2

Thus,

gh = \frac{v_1^2-v_2^2}{2} + C_p(T_1-T_2)

Now,

C_p = \frac{\gamma R}{\gamma-1}

We first find $R$ using ideal gas law at lower end:

P_1 = \rho_1 R T_1

R = \frac{P_1}{\rho_1 T_1} = \frac{600}{0.2\times 300} = \frac{600}{60}=10\,\text{J kg}^{-1}\text{K}^{-1}

Hence,

C_p = \frac{2\times 10}{2-1} = 20\,\text{J kg}^{-1}\text{K}^{-1}

Now calculate:

\frac{v_1^2-v_2^2}{2} = \frac{40^2-20^2}{2} = \frac{1600-400}{2} = 600

C_p(T_1-T_2)=20(300-150)=20\times 150=3000

Therefore,

gh = 600+3000=3600

h = \frac{3600}{10}=360\,\text{m}

So statement C is false.

Density at upper end

We already found:

\rho_2 = 0.1\,\text{kg m}^{-3}

So statement D is false.

Final evaluation of options

A: False
B: True
C: False
D: False

Hence, the correct option is:

\boxed{\text{B only}}

Comparison with stored correct answer

Stored correct answer: B

My derived answer: B only

They agree.

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