Question

In a cylindrical container of sufficiently large height, two easily moving pistons enclose certain amount of same ideal gas in two chambers as shown in the figure-2.43.

The upper piston is at a height $20\, cm$ from the bottom and lower piston is at a height $8\, cm$ from the bottom. The mass of each piston is m kg and cross sectional area of each piston is $A\, m^2$ where $\frac{m g}{A}=P_{0}$ and $P_{0}$ is the atmospheric pressure $=1 \times 10^{5} N / m ^{2}$.
The cylindrical container and pistons are made of conducting material. Initially the temperature of gas is $27^{\circ} C$ and whole system is in equilibrium. Now if the upper piston is slowly lifted by $16\, cm$ and held in that position with the help of some external force. As a result, the lower piston rises slowly by $l\, cm$. In a cylindrical container of sufficiently large height, two easily moving pistons enclose certain amount of same ideal gas in two chambers as shown in the figure-2.43.

The upper piston is at a height $20\, cm$ from the bottom and lower piston is at a height $8\, cm$ from the bottom. The mass of each piston is m kg and cross sectional area of each piston is $A\, m^2$ where $\frac{m g}{A}=P_{0}$ and $P_{0}$ is the atmospheric pressure $=1 \times 10^{5} N / m ^{2}$.
The cylindrical container and pistons are made of conducting material. Initially the temperature of gas is $27^{\circ} C$ and whole system is in equilibrium. Now if the upper piston is slowly lifted by $16\, cm$ and held in that position with the help of some external force. As a result, the lower piston rises slowly by $l\, cm$.
The value of $l$ is :

• A. 2 cm
• B. 4 cm
• C. 8 cm
• D. 6 cm

Answer 1

Answer: (B)

Let $P_{1}$ and $P_{2}$ be the initial pressure in lower chamber of gas and upper chamber of gas.
$P_{2}=P_{0}+\frac{m g}{A}=2 P_{0}, V_{2}=A \times 12 \times 10^{-2} m ^{3}$
If $P_{2}^{\prime}$ and $V_{2}^{\prime}$ are final pressure and volume in upper chamber
$V_{2}^{\prime} =A \times(28-l) \times 10^{-2} m ^{3}$
$P_{2} V_{2} =P_{2}^{\prime} V_{2}^{\prime}$
$\Rightarrow P_{2}^{\prime}=\frac{P_{2} V_{2}}{P_{2}^{\prime}}=\frac{24 P_{0}}{28-l}$
Now consider lower chamber
$P_{1} =P_{0}+\frac{2 m g}{A}=3 P_{0}$
and $V_{1} =A \times 8 \times 10^{-2} m ^{3}$
$P_{1}^{\prime} =P_{2}^{\prime}+\frac{m g}{A}=P_{0}\left[\frac{52-l}{28-l}\right]$
and $V_{1}^{\prime} =A \times(8+l) \times 10^{-2} m ^{3}$
$P_{1} V_{1} =P_{1}^{\prime} V_{1}^{\prime}$
$3 P_{0} A \times 8 \times 10^{-2} =P_{0}\left[\frac{52-l}{28-l}\right] \times A \times(8+l) \times 10^{-2}$
$24 =\left[\frac{52-l}{28-l}\right] \times(8+l)$
Solving we get $l=4\, cm$
$\Rightarrow P_{1}^{\prime}=2 P_{0}=2 \times 10^{5} A / m ^{2}$
$\Rightarrow P_{2}^{\prime}=\frac{24 P_{0}}{28-l}=P_{0}=1 \times 10^{5} N / m ^{2}$
$\frac{V_{2}^{\prime}}{V_{1}^{\prime}}=\frac{28-l}{8+l}=\frac{24}{12}=2$