JEE Advance - Physics (2022 - Paper 1 Online - No. 11)
An ideal gas of density $\rho=0.2 \mathrm{~kg} \mathrm{~m}^{-3}$ enters a chimney of height $h$ at the rate of $\alpha=$ $0.8 \mathrm{~kg} \mathrm{~s}^{-1}$ from its lower end, and escapes through the upper end as shown in the figure. The cross-sectional area of the lower end is $A_{1}=0.1 \mathrm{~m}^{2}$ and the upper end is $A_{2}=0.4 \mathrm{~m}^{2}$. The pressure and the temperature of the gas at the lower end are $600 \mathrm{~Pa}$ and $300 \mathrm{~K}$, respectively, while its temperature at the upper end is $150 \mathrm{~K}$. The chimney is heat insulated so that the gas undergoes adiabatic expansion. Take $g=10 \mathrm{~m} \mathrm{~s}^{-2}$ and the ratio of specific heats of the gas $\gamma=2$. Ignore atmospheric pressure.
Which of the following statement(s) is(are) correct?
The pressure of the gas at the upper end of the chimney is $300 \mathrm{~Pa}$.
The velocity of the gas at the lower end of the chimney is $40 \mathrm{~m} \mathrm{~s}^{-1}$ and at the upper end is $20 \mathrm{~ms}^{-1}$.
The height of the chimney is $590 \mathrm{~m}$.
The density of the gas at the upper end is $0.05 \mathrm{~kg} \mathrm{~m}^{-3}$.
Explanation
Given, $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2: \gamma=2$
The process described is adiabatic. In an adiabatic process (a process during which no heat is gained or lost), the following relationship holds:
$P_1^{1-\gamma} \cdot T_1^\gamma = P_2^{1-\gamma} \cdot T_2^\gamma$
Where :
- $P_1$ and $P_2$ are the initial and final pressures,
- $T_1$ and $T_2$ are the initial and final temperatures,
- $\gamma$ is the ratio of specific heats (given as 2).
Given that $P_1 = 600 \, \text{Pa}$, $T_1 = 300 \, \text{K}$, $T_2 = 150 \, \text{K}$, and $\gamma = 2$, solve for $P_2$ :
$\frac{P_2}{P_1} = \left(\frac{T_1}{T_2}\right)^{\gamma/(1-\gamma)}$
$\Rightarrow P_2 = P_1 \left(\frac{T_1}{T_2}\right)^{\gamma/(1-\gamma)}$
$\Rightarrow P_2 = 600 \, \text{Pa} \times \left(\frac{300 \, \text{K}}{150 \, \text{K}}\right)^{2/(1-2)}$
$\Rightarrow P_2 = 150 \, \text{Pa}$
Using the ideal gas law in the form $\rho = \frac{PM}{RT}$ (where R is the specific gas constant), and rearranging to find the relationship between $\rho_1$ and $\rho_2$ :
$\frac{\rho_1}{\rho_2} = \frac{P_1}{P_2} \cdot \frac{T_2}{T_1}$
$\Rightarrow \rho_2 = \rho_1 \cdot \frac{P_2}{P_1} \cdot \frac{T_1}{T_2}$
$\Rightarrow \rho_2 = 0.2 \, \text{kg/m}^3 \cdot \frac{150 \, \text{Pa}}{600 \, \text{Pa}} \cdot \frac{300 \, \text{K}}{150 \, \text{K}}$
$\Rightarrow \rho_2 = 0.1 \, \text{kg/m} ^3$
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, $\frac{d m}{d t} = \rho_1 \cdot A_1 \cdot v_1 = \rho_2 \cdot A_2 \cdot v_2$,
The mass flow rate is conserved, and given by $\alpha = \rho \cdot A \cdot v$. Thus, you have:
$\alpha = \rho_1 \cdot A_1 \cdot v_1$
$\Rightarrow 0.8 \, \text{kg/s} = 0.2 \, \text{kg/m}^3 \cdot 0.1 \, \text{m}^2 \cdot v_1$
$\Rightarrow v_1 = \frac{0.8 \, \text{kg/s}}{0.2 \, \text{kg/m}^3 \cdot 0.1 \, \text{m}^2} = 40 \, \text{m/s}$
Similarly, for the upper end :
$\alpha = \rho_2 \cdot A_2 \cdot v_2$
$\Rightarrow v_2 = \frac{0.8 \, \text{kg/s}}{0.1 \, \text{kg/m}^3 \cdot 0.4 \, \text{m}^2} = 20 \, \text{m/s}$
Work done on the gas $=$ total energy $=\Delta \mathrm{K}+\Delta \mathrm{U}+$ internal energy
$$ \begin{aligned} & \mathrm{P}_1 \mathrm{~A}_1 \Delta x_1-\mathrm{P}_2 \mathrm{~A}_2 \Delta x_2=\frac{1}{2} \Delta m v_2^2-\frac{1}{2} \Delta m v_1^2+\Delta m g h+\frac{f}{2} \\\\ &\left(\mathrm{P}_2 \Delta \mathrm{V}_2-\mathrm{P}_1 \Delta \mathrm{V}_1\right) 2 \mathrm{P}_1 \frac{\Delta \mathrm{v}_1}{\Delta m}-2 \mathrm{P}_2 \frac{\Delta \mathrm{v}_2}{\Delta m}=\frac{\mathrm{v}_2^2-\mathrm{v}_1^2}{2}+g h \\\\ & \frac{2 \times 600}{0.2}-\frac{2 \times 150}{0.1}=\frac{20^2-40^2}{2}+10 \mathrm{~h} \\\\ & 6000-3000=\frac{400-1600}{2}+10 \mathrm{~h} \\\\ & 3000=-600+10 \mathrm{~h} \\\\ & 10 \mathrm{~h}=3000+600 \\\\ & \mathrm{~h}=360 \mathrm{~m} \end{aligned} $$
The process described is adiabatic. In an adiabatic process (a process during which no heat is gained or lost), the following relationship holds:
$P_1^{1-\gamma} \cdot T_1^\gamma = P_2^{1-\gamma} \cdot T_2^\gamma$
Where :
- $P_1$ and $P_2$ are the initial and final pressures,
- $T_1$ and $T_2$ are the initial and final temperatures,
- $\gamma$ is the ratio of specific heats (given as 2).
Given that $P_1 = 600 \, \text{Pa}$, $T_1 = 300 \, \text{K}$, $T_2 = 150 \, \text{K}$, and $\gamma = 2$, solve for $P_2$ :
$\frac{P_2}{P_1} = \left(\frac{T_1}{T_2}\right)^{\gamma/(1-\gamma)}$
$\Rightarrow P_2 = P_1 \left(\frac{T_1}{T_2}\right)^{\gamma/(1-\gamma)}$
$\Rightarrow P_2 = 600 \, \text{Pa} \times \left(\frac{300 \, \text{K}}{150 \, \text{K}}\right)^{2/(1-2)}$
$\Rightarrow P_2 = 150 \, \text{Pa}$
Using the ideal gas law in the form $\rho = \frac{PM}{RT}$ (where R is the specific gas constant), and rearranging to find the relationship between $\rho_1$ and $\rho_2$ :
$\frac{\rho_1}{\rho_2} = \frac{P_1}{P_2} \cdot \frac{T_2}{T_1}$
$\Rightarrow \rho_2 = \rho_1 \cdot \frac{P_2}{P_1} \cdot \frac{T_1}{T_2}$
$\Rightarrow \rho_2 = 0.2 \, \text{kg/m}^3 \cdot \frac{150 \, \text{Pa}}{600 \, \text{Pa}} \cdot \frac{300 \, \text{K}}{150 \, \text{K}}$
$\Rightarrow \rho_2 = 0.1 \, \text{kg/m} ^3$
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, $\frac{d m}{d t} = \rho_1 \cdot A_1 \cdot v_1 = \rho_2 \cdot A_2 \cdot v_2$,
The mass flow rate is conserved, and given by $\alpha = \rho \cdot A \cdot v$. Thus, you have:
$\alpha = \rho_1 \cdot A_1 \cdot v_1$
$\Rightarrow 0.8 \, \text{kg/s} = 0.2 \, \text{kg/m}^3 \cdot 0.1 \, \text{m}^2 \cdot v_1$
$\Rightarrow v_1 = \frac{0.8 \, \text{kg/s}}{0.2 \, \text{kg/m}^3 \cdot 0.1 \, \text{m}^2} = 40 \, \text{m/s}$
Similarly, for the upper end :
$\alpha = \rho_2 \cdot A_2 \cdot v_2$
$\Rightarrow v_2 = \frac{0.8 \, \text{kg/s}}{0.1 \, \text{kg/m}^3 \cdot 0.4 \, \text{m}^2} = 20 \, \text{m/s}$
Work done on the gas $=$ total energy $=\Delta \mathrm{K}+\Delta \mathrm{U}+$ internal energy
$$ \begin{aligned} & \mathrm{P}_1 \mathrm{~A}_1 \Delta x_1-\mathrm{P}_2 \mathrm{~A}_2 \Delta x_2=\frac{1}{2} \Delta m v_2^2-\frac{1}{2} \Delta m v_1^2+\Delta m g h+\frac{f}{2} \\\\ &\left(\mathrm{P}_2 \Delta \mathrm{V}_2-\mathrm{P}_1 \Delta \mathrm{V}_1\right) 2 \mathrm{P}_1 \frac{\Delta \mathrm{v}_1}{\Delta m}-2 \mathrm{P}_2 \frac{\Delta \mathrm{v}_2}{\Delta m}=\frac{\mathrm{v}_2^2-\mathrm{v}_1^2}{2}+g h \\\\ & \frac{2 \times 600}{0.2}-\frac{2 \times 150}{0.1}=\frac{20^2-40^2}{2}+10 \mathrm{~h} \\\\ & 6000-3000=\frac{400-1600}{2}+10 \mathrm{~h} \\\\ & 3000=-600+10 \mathrm{~h} \\\\ & 10 \mathrm{~h}=3000+600 \\\\ & \mathrm{~h}=360 \mathrm{~m} \end{aligned} $$
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