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ssm A cylindrical glass beaker of height 1.520 \(\mathrm{m}\) rests on a table. The bottom half of the beaker is filled with a gas, and the top half is filled with liquid mercury that is exposed to the atmosphere. The gas and mercury do not mix because they are separated by a frictionless movable piston of negligible mass and thickness. The initial temperature is 273 \(\mathrm{K}\) . The temperature is increased until a value is reached when one-half of the mercury has spilled out. Ignore the thermal expansion of the glass and the mercury, and find this temperature.

Short Answer

Expert verified
The final temperature is approximately 409 K.

Step by step solution

01

Define the System

We have a cylindrical beaker with a gas and mercury separated by a movable piston. The bottom half contains a gas and the top half is filled with liquid mercury. The system is initially at 273 K.
02

Note Important Constants

The initial height of gas under the piston is half the beaker's height: \(h_0 = \frac{1.520}{2} \text{ m} = 0.760 \text{ m}\). The initial temperature \(T_0 = 273 \text{ K}\).
03

Understand Volume-Temperature Relation for the Gas

According to Charles's law, the volume of a gas varies directly with temperature, assuming pressure remains constant. The relation can be given as \( \frac{V_1}{T_0} = \frac{V_2}{T_2} \), where \(V_1\) and \(V_2\) are initial and final volumes, \(T_0\) and \(T_2\) are initial and final temperatures.
04

Calculate Initial Volume of the Gas

The initial volume \(V_1\) of the gas is given by \(V_1 = \pi r^2 h_0\), and since \(h_0=0.760 \text{ m}\), \(V_1 = \pi r^2 \times 0.760\). However, we will see that \(r^2\) cancels out in the next steps, so no need to plug in values here.
05

Evaluate Final State of the System

When half of the mercury is spilled, the new height of mercury is \(\frac{0.760}{2} \text{ m} = 0.380 \text{ m}\). Hence, the final height of the gas is then \(0.760 \text{ m} + 0.380 \text{ m} = 1.140 \text{ m}\).
06

Calculate Final Volume of the Gas

The final volume \(V_2\) of the gas is \(V_2 = \pi r^2 \times 1.140\).
07

Apply Charles’s Law and Solve for Final Temperature

Using \( \frac{V_1}{T_0} = \frac{V_2}{T_2} \), substitute the volumes:\[\frac{\pi r^2 \times 0.760 \, \text{m}}{273 \, \text{K}} = \frac{\pi r^2 \times 1.140 \, \text{m}}{T_2}\]\[T_2 = 273 \times \frac{1.140}{0.760} \approx 409 \, \text{K}\].

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Key Concepts

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

Thermal Expansion
Thermal expansion is the increase in volume of a substance due to an increase in temperature. In many contexts, like this problem, we assume the container's expansion is negligible. Here, the glass beaker doesn't change, simplifying the calculations of volume changes for the enclosed gas. If the container expanded, we might have to reconsider the arrangement and volume under increasing temperatures. Charles’s law helps us understand thermal expansion in gases by stating that the volume of a gas is directly proportional to its temperature when pressure remains constant. This means as temperature increases, the gas volume will increase. Though crucially, in this exercise, mercury's thermal expansion is ignored, focusing on the gas expanding within the beaker instead.
Cylindrical Beaker
A cylindrical beaker serves as a simple model that facilitates calculations of volume and height. It is an important aspect since it affects how the enclosed gas behaves when temperatures change. The shape of the beaker ensures that the volume can be directly related to its height by the formula for the volume of a cylinder: \(V = \pi r^2 h\). This symmetry simplifies how we determine the changes in height and volume of the gas as the temperature varies. In this particular problem, understanding that the beaker is a cylinder helps isolate variables and apply Charles's law to find desired temperatures. When one half of the mercury spills out, the gas can expand further: a straightforward result of operating within this symmetrical shape.
Movable Piston
The movable piston is a fundamental part of this problem because it allows the gas and liquid mercury to exist in non-mixing sections of the beaker. It is not influenced by friction or mass, which makes it a perfect barrier between the two phases. This piston adjusts its position automatically as the gas expands due to heating, allowing the mercury to be displaced. If the piston had friction, it might resist motion, therefore altering expected results. The piston's role is crucial in demonstrating how the gas obeys Charles’s law without interference from other forces. As one half of the mercury is displaced, only the gas's expansion affects the system dynamics, highlighting the piston's importance in our calculations.
Initial and Final Temperatures
The initial and final temperatures define the conditions that dictate the gas's behavior under Charles’s law. Initially, the temperature is 273 K, at which point the gas occupies half the volume of the beaker. When the temperature increases, so does the gas volume, which pushes up against the piston and displaces the mercury above it. The problem focuses on finding the temperature at which half of the mercury spills out. The relationship between volume and temperature remains a constant factor. Through computations, we determine that the final temperature, when the desired displacement occurs, is approximately 409 K. This precise end condition solidifies the theoretical predictions based on known gas laws and conditions. Understanding these temperatures allows us to visualize and calculate the gas's response to heating within the confined system.

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