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QJEEAdv22Phe860: JEE Advanced 2022 Physics - Class 11 JEEAdv22Phe860 Solution: An ideal gas of density enters a chimney of height at the rate…

An ideal gas of density ρ=0.2 kg m−3\rho=0.2 \mathrm{~kg} \mathrm{~m}^{-3} enters a chimney of height hh at the rate of α=\alpha= 0.8 kg s−10.8 \mathrm{~kg} \mathrm{~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 A1=0.1 m2A_{1}=0.1 \mathrm{~m}^{2} and the upper end is A2=0.4 m2A_{2}=0.4 \mathrm{~m}^{2} . The pressure and the temperature of the gas at the lower end are 600 Pa600 \mathrm{~Pa} and 300 K300 \mathrm{~K} , respectively, while its temperature at the upper end is 150 K150 \mathrm{~K} . The chimney is heat insulated so that the gas undergoes adiabatic expansion. Take g=10 m s−2g=10 \mathrm{~m} \mathrm{~s}^{-2} and the ratio of specific heats of the gas γ=2\gamma=2 . Ignore atmospheric pressure.

JEE Advanced 2022 Paper 1 Online Physics - Properties of Matter Question 14 English

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

Multiple Choice Options

  • A The pressure of the gas at the upper end of the chimney is 300 Pa300 \mathrm{~Pa} .
  • B The velocity of the gas at the lower end of the chimney is 40 m s−140 \mathrm{~m} \mathrm{~s}^{-1} and at the upper end is 20 ms−120 \mathrm{~ms}^{-1} .
  • C The height of the chimney is 590 m590 \mathrm{~m} .
  • D The density of the gas at the upper end is 0.05 kg m−30.05 \mathrm{~kg} \mathrm{~m}^{-3} .

Step-by-Step Verified Solution

Given, g=10 m/s2:γ=2\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2: \gamma=2

The process described is adiabatic. In an adiabatic process (a process during which no heat is gained or lost), the following relationship holds:

P11−γ⋅T1γ=P21−γ⋅T2γP_1^{1-\gamma} \cdot T_1^\gamma = P_2^{1-\gamma} \cdot T_2^\gamma

Where :

- P1P_1 and P2P_2 are the initial and final pressures,

- T1T_1 and T2T_2 are the initial and final temperatures,

- γ\gamma is the ratio of specific heats (given as 2).

Given that P1=600 PaP_1 = 600 \, \text{Pa} , T1=300 KT_1 = 300 \, \text{K} , T2=150 KT_2 = 150 \, \text{K} , and γ=2\gamma = 2 , solve for P2P_2 :

P2P1=(T1T2)γ/(1−γ)\frac{P_2}{P_1} = \left(\frac{T_1}{T_2}\right)^{\gamma/(1-\gamma)}

⇒P2=P1(T1T2)γ/(1−γ)\Rightarrow P_2 = P_1 \left(\frac{T_1}{T_2}\right)^{\gamma/(1-\gamma)}

⇒P2=600 Pa×(300 K150 K)2/(1−2)\Rightarrow P_2 = 600 \, \text{Pa} \times \left(\frac{300 \, \text{K}}{150 \, \text{K}}\right)^{2/(1-2)}

⇒P2=150 Pa\Rightarrow P_2 = 150 \, \text{Pa}

Using the ideal gas law in the form ρ=PMRT\rho = \frac{PM}{RT} (where R is the specific gas constant), and rearranging to find the relationship between ρ1\rho_1 and ρ2\rho_2 :

ρ1ρ2=P1P2⋅T2T1\frac{\rho_1}{\rho_2} = \frac{P_1}{P_2} \cdot \frac{T_2}{T_1}

⇒ρ2=ρ1⋅P2P1⋅T1T2\Rightarrow \rho_2 = \rho_1 \cdot \frac{P_2}{P_1} \cdot \frac{T_1}{T_2}

⇒ρ2=0.2 kg/m3⋅150 Pa600 Pa⋅300 K150 K\Rightarrow \rho_2 = 0.2 \, \text{kg/m}^3 \cdot \frac{150 \, \text{Pa}}{600 \, \text{Pa}} \cdot \frac{300 \, \text{K}}{150 \, \text{K}}

⇒ρ2=0.1 kg/m3\Rightarrow \rho_2 = 0.1 \, \text{kg/m}^3

Next, use the principle of conservation of mass flow rate to determine the velocity of the gas at the lower and upper ends.

Mass flow rate, dmdt=ρ1⋅A1⋅v1=ρ2⋅A2⋅v2\frac{d m}{d t} = \rho_1 \cdot A_1 \cdot v_1 = \rho_2 \cdot A_2 \cdot v_2 ,

The mass flow rate is conserved, and given by α=ρ⋅A⋅v\alpha = \rho \cdot A \cdot v . Thus, you have:

α=ρ1⋅A1⋅v1\alpha = \rho_1 \cdot A_1 \cdot v_1

⇒0.8 kg/s=0.2 kg/m3⋅0.1 m2⋅v1\Rightarrow 0.8 \, \text{kg/s} = 0.2 \, \text{kg/m}^3 \cdot 0.1 \, \text{m}^2 \cdot v_1

⇒v1=0.8 kg/s0.2 kg/m3⋅0.1 m2=40 m/s\Rightarrow v_1 = \frac{0.8 \, \text{kg/s}}{0.2 \, \text{kg/m}^3 \cdot 0.1 \, \text{m}^2} = 40 \, \text{m/s}

Similarly, for the upper end :

α=ρ2⋅A2⋅v2\alpha = \rho_2 \cdot A_2 \cdot v_2

⇒v2=0.8 kg/s0.1 kg/m3⋅0.4 m2=20 m/s\Rightarrow v_2 = \frac{0.8 \, \text{kg/s}}{0.1 \, \text{kg/m}^3 \cdot 0.4 \, \text{m}^2} = 20 \, \text{m/s}

Work done on the gas == total energy =ΔK+ΔU+=\Delta \mathrm{K}+\Delta \mathrm{U}+ internal energy

amp;P1 A1Δx1−P2 A2Δx2=12Δmv22−12Δmv12+Δmgh+f2amp;(P2ΔV2−P1ΔV1)2P1Δv1Δm−2P2Δv2Δm=v22−v122+ghamp;2×6000.2−2×1500.1=202−4022+10 hamp;6000−3000=400−16002+10 hamp;3000=−600+10 hamp;10 h=3000+600amp; h=360 m\begin{aligned}& \mathrm{P}_1 \mathrm{~A}_1 \Delta x_1-\mathrm{P}_2 \mathrm{~A}_2 \Delta x_2=\frac{1}{2} \Delta m v_2^2-\frac{1}{2} \Delta m v_1^2+\Delta m g h+\frac{f}{2} \\\\&\left(\mathrm{P}_2 \Delta \mathrm{V}_2-\mathrm{P}_1 \Delta \mathrm{V}_1\right) 2 \mathrm{P}_1 \frac{\Delta \mathrm{v}_1}{\Delta m}-2 \mathrm{P}_2 \frac{\Delta \mathrm{v}_2}{\Delta m}=\frac{\mathrm{v}_2^2-\mathrm{v}_1^2}{2}+g h \\\\& \frac{2 \times 600}{0.2}-\frac{2 \times 150}{0.1}=\frac{20^2-40^2}{2}+10 \mathrm{~h} \\\\& 6000-3000=\frac{400-1600}{2}+10 \mathrm{~h} \\\\& 3000=-600+10 \mathrm{~h} \\\\& 10 \mathrm{~h}=3000+600 \\\\& \mathrm{~h}=360 \mathrm{~m}\end{aligned}

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