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JEE Advanced 2023 · Thermodynamics

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

Diagram for this Thermodynamics question (JEE Advanced 2023).

Answers: B. The velocity of the gas at the lower end of the chimney is 40 m s−140\,\text{m s}^{-1} and at the upper end is 20 m s−120\,\text{m s}^{-1}.

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Written solution

Step 1: Calculate the velocity of the gas at the lower end.

The mass flow rate is constant throughout the chimney.

α=ρ1A1v1\alpha = \rho_1 A_1 v_1 v1=αρ1A1v_1 = \frac{\alpha}{\rho_1 A_1} v1=0.80.2×0.1=40 m s−1v_1 = \frac{0.8}{0.2 \times 0.1} = 40\,\text{m s}^{-1}

Step 2: Determine the density of the gas at the upper end.

For an adiabatic process of an ideal gas, the relation between temperature and density is given by the adiabatic relation.

Tρ1−γ=constantT \rho^{1-\gamma} = \text{constant} ρ∝T1γ−1\rho \propto T^{\frac{1}{\gamma - 1}}

Substitute γ=2\gamma = 2.

ρ2=ρ1(T2T1)\rho_2 = \rho_1 \left(\frac{T_2}{T_1}\right) ρ2=0.2×150300=0.1 kg m−3\rho_2 = 0.2 \times \frac{150}{300} = 0.1\,\text{kg m}^{-3}

Step 3: Calculate the velocity of the gas at the upper end.

Apply the mass flow rate equation at the upper end.

α=ρ2A2v2\alpha = \rho_2 A_2 v_2 v2=αρ2A2v_2 = \frac{\alpha}{\rho_2 A_2} v2=0.80.1×0.4=20 m s−1v_2 = \frac{0.8}{0.1 \times 0.4} = 20\,\text{m s}^{-1}

Step 4: Calculate the pressure of the gas at the upper end.

Use the ideal gas equation of state.

pρT=constant\frac{p}{\rho T} = \text{constant} p2=p1(ρ2T2ρ1T1)p_2 = p_1 \left(\frac{\rho_2 T_2}{\rho_1 T_1}\right) p2=600×(0.1×1500.2×300)=150 Pap_2 = 600 \times \left(\frac{0.1 \times 150}{0.2 \times 300}\right) = 150\,\text{Pa}

Step 5: Calculate the height of the chimney.

Apply the energy conservation equation for steady adiabatic flow.

v22+gh+γγ−1pρ=constant\frac{v^2}{2} + gh + \frac{\gamma}{\gamma-1}\frac{p}{\rho} = \text{constant} v122+γγ−1p1ρ1=v222+gh+γγ−1p2ρ2\frac{v_1^2}{2} + \frac{\gamma}{\gamma-1}\frac{p_1}{\rho_1} = \frac{v_2^2}{2} + gh + \frac{\gamma}{\gamma-1}\frac{p_2}{\rho_2} h=12(v12−v22)+γγ−1(p1ρ1−p2ρ2)gh = \frac{\frac{1}{2}(v_1^2 - v_2^2) + \frac{\gamma}{\gamma-1}\left(\frac{p_1}{\rho_1} - \frac{p_2}{\rho_2}\right)}{g} h=12(402−202)+22−1(6000.2−1500.1)10h = \frac{\frac{1}{2}(40^2 - 20^2) + \frac{2}{2-1}\left(\frac{600}{0.2} - \frac{150}{0.1}\right)}{10} h=600+2(3000−1500)10h = \frac{600 + 2(3000 - 1500)}{10} h=600+300010=360 mh = \frac{600 + 3000}{10} = 360\,\text{m}

Answer: The correct statement is that the velocity of the gas at the lower end is v1=40 m s−1v_1 = 40\,\text{m s}^{-1} and at the upper end is v2=20 m s−1v_2 = 20\,\text{m s}^{-1}.

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