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Chapter 24: Problem 30

A vertical cylinder of height \(100 \mathrm{~cm}\) contains air at a constant temperature. The top is closed by a frictionless light piston. The atmospheric pressure is equal to \(75 \mathrm{~cm}\) of mercury. Mercury is slowly poured over the piston. Find the maximum height of the mercury column that can be put on the piston.

Short Answer

Expert verified
The maximum height of the mercury column is 25 cm.

Step by step solution

01

Identify the Known Values

We are given a cylinder with height \( h_c = 100 \text{ cm} \). The atmospheric pressure \( P_0 \) is equivalent to \( 75 \text{ cm of Hg} \). Our task is to find the maximum height of the mercury column, \( h_m \), that can be added to the piston without compressing the air inside beyond the height of the cylinder.
02

Define Pressure Conditions

The air inside the cylinder initially experiences atmospheric pressure, so the initial pressure, \( P_i \), is equal to \( 75 \text{ cm of Hg} \). When the piston is covered with mercury, the final pressure \( P_f \) will be the sum of the atmospheric pressure and the pressure due to the mercury column: \( P_f = 75 + h_m \), where \( h_m \) is the height of the mercury column.
03

Apply Boyle's Law

Since the temperature is constant, we apply Boyle's Law: \( P_iV_i = P_fV_f \). Here, \( V_i = A \times h_c \) and \( V_f = A \times (h_c - h_m) \), where \( A \) is the cross-sectional area of the cylinder. This simplifies to: \( 75 \times 100 = (75 + h_m) \times (100 - h_m) \).
04

Simplify and Solve for $h_m$

Using the equation from Step 3: \[ 7500 = (75 + h_m)(100 - h_m) \]Expand the equation:\[ 7500 = 7500 - 75h_m + 100h_m - h_m^2 \]This simplifies to:\[ 0 = 25h_m - h_m^2 \]Factoring gives us:\[ h_m(h_m - 25) = 0 \]Thus, \( h_m = 0 \) or \( h_m = 25 \text{ cm} \).
05

Interpret the Result

Since \( h_m = 0 \) corresponds to no mercury added and doesn't provide the maximum possible height, \( h_m = 25 \text{ cm} \) is our valid solution. Therefore, the maximum height of the mercury column is \( 25 \text{ cm} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pressure Conditions
Understanding the pressure conditions in this scenario is essential for solving the problem. In our exercise, we are dealing with a cylinder filled with air and a piston on top. The air initially experiences the atmospheric pressure, which is given in the problem as equivalent to 75 cm of mercury (Hg).
When mercury is poured over the piston, it adds its weight as additional pressure inside the cylinder.
This creates a new pressure condition, where the final pressure, \( P_f \), is now the sum of the atmospheric pressure and the pressure from the mercury column.
  • Initial Pressure \( P_i \): 75 cm of Hg
  • Final Pressure \( P_f \): \( 75 + h_m \)
Where \( h_m \) is the height of the mercury column. Understanding these pressure changes is crucial for applying Boyle's Law in the exercise.
Cylinder Volume
The volume of the cylinder is a key concept when applying Boyle鈥檚 Law. In this problem, the volume of the air changes as the mercury is added. Initially, the air fills the cylinder completely, so its volume \( V_i \) is given by the product of the cross-sectional area \( A \) and the height of the cylinder \( h_c \), which is 100 cm.
However, as mercury is added, the volume of the air is compressed because the piston moves downward, making the final volume \( V_f \)
  • Initial Volume \( V_i = A \times 100 \)
  • Final Volume \( V_f = A \times (100 - h_m) \)
When using Boyle鈥檚 Law, these changes in volume allow us to relate the initial and final conditions through the pressure-volume relationship \( P_iV_i = P_fV_f \). Due to the constancy of temperature, Boyle's Law assumes that the product of pressure and volume remains the same.
Mercury Column Pressure
The pressure exerted by the mercury column plays a crucial role in resolving the exercise. When mercury is added over the piston, it contributes additional pressure due to its weight. This additional pressure is described as the height of the mercury column, \( h_m \), measured in centimeters of mercury (cm of Hg).
The final pressure inside the cylinder is influenced by both the atmospheric pressure and this new pressure from the mercury column. This is why we calculate the final pressure \( P_f \) by adding the atmospheric pressure and the height of the mercury column:
  • Final Pressure \( P_f = 75 + h_m \)
The height of the mercury column can't exceed a certain value without compressing the air in the cylinder. Solving the exercise calculated that the maximum height \( h_m \) allowed is 25 cm. This value ensures that the air remains at a pressure that the volume change through Boyle's Law can accommodate.

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Most popular questions from this chapter

The temperature and the relative humidity are \(300 \mathrm{~K}\) and \(20 \%\) in a room of volume \(50 \mathrm{~m}^{3}\), The floor is washed with water, \(500 \mathrm{~g}\) of water sticking on the floor. Assuming no communication with the surrounding, find the relative humidity when the floor dries. The changes in temperature and pressure may be neglected. Saturation vapour pressure at \(300 \mathrm{~K}=3.3 \mathrm{kPa}\).

A uniform tube closed at one end, contains a pellet of mercury \(10 \mathrm{~cm}\) long. When the tube is kept vertically with the closed-end upward, the length of the air column trapped is \(20 \mathrm{~cm}\). Find the length of the air column trapped when the tube is inverted so that the closedend goes down. Atmospheric pressure \(=75 \mathrm{~cm}\) of mercury.

A vessel of volume \(V_{0}\) contains an ideal gas at pressure \(p_{0}\) and temperature \(T\). Gas is continuously pumped out of this vessel at a constant volume-rate \(d V / d t=r\) keeping the temperature constant. The pressure of the gas being taken out equals the pressure inside the vessel. Find (a) the pressure of the gas as a function of time, (b) the time taken before half the original gas is pumped out.

The temperature and humidity of air are \(27^{\circ} \mathrm{C}\) and \(50 \%\) on a particular day. Calculate the amount of vapour that should be added to 1 cubic metre of air to saturate it. The saturation vapour pressure at \(27^{\circ} \mathrm{C}=3600 \mathrm{~Pa}\).

An electric bulb of volume \(250 \mathrm{cc}\) was sealed during manufacturing at a pressure of \(10^{-3} \mathrm{~mm}\) of mercury at \(27^{\circ} \mathrm{C}\). Compute the number of air molecules contained in the bulb. Avogadro constant \(=6 \times 10^{23} \mathrm{~mol}^{-1}\), density of mercury \(=13600 \mathrm{~kg} \mathrm{~m}^{-3}\) and \(g=10 \mathrm{~m} \mathrm{~s}^{-2}\).
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