Problem 27 - Physics

Question

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 mercury, and find this temperature.

Step-by-Step Solution

Answer

The temperature is approximately 409.5 K.

Step 1: Understand Initial Conditions

Initially, the beaker is composed of half gas and half mercury. Consider the beaker's height to be evenly split: 0.760 m of gas and 0.760 m of mercury.

Step 2: Apply Ideal Gas Law to Initial State

The initial state can be described by the ideal gas law, \( PV = nRT \), where:- \( P \) is the pressure (1 atm since it's atmospheric pressure), - \( V \) is the initial volume occupied by gas (\( 0.760 \times A \), where \( A \) is the cross-sectional area),- \( n \) is the number of moles,- \( R \) is the ideal gas constant,- \( T \) is the initial temperature (273 K).

Step 3: Analyze Conditions After Mercury Spills

After raising the temperature, half the mercury is spilled, reducing its height to 0.380 m, allowing the gas to expand to a height of 1.140 m. The pressure on the gas is still 1 atm due to atmospheric exposure.

Step 4: Apply Ideal Gas Law to Final State

For the final state when half the mercury is spilled out, the equation becomes \( P'V' = nRT' \). Here,- \( P' = 1\, \text{atm} \),- \( V' = 1.140 \times A \),- \( T' \) is the unknown temperature we need to find.

Step 5: Set Up the Equation for Gas Expansion

Since the number of moles of gas and pressure remain constant, the equations become:\[ \frac{P_i V_i}{T_i} = \frac{P_f V_f}{T_f} \]This simplifies to:\[ \frac{V_i}{T_i} = \frac{V_f}{T_f} \] \ With \( V_i = 0.760 \times A \) and \( V_f = 1.140 \times A \).

Step 6: Solve for Unknown Temperature

Now solve the equation:\[ \frac{0.760}{273} = \frac{1.140}{T'} \]Cross-multiply and solve for \( T' \):\[ T' = \frac{1.140 \times 273}{0.760} \approx 409.5 \text{ K} \]

Key Concepts

Cylindrical Beaker

A cylindrical beaker is a container with a circular cross-section and a specific height, which in this exercise measures to 1.520 meters. The design of the beaker is important because it allows us to assume uniform distribution of materials along its height.
This shape also facilitates the application of physical laws, such as the Ideal Gas Law, because calculations involving volume become straightforward. In this problem, the beaker is initially half-filled with gas and half with liquid mercury.

• The cylindrical shape ensures that the volume is proportional to the height, simplifying calculation of volumes for both gas and mercury sections.
• The cross-sectional area (\( A \)) plays a crucial role in determining the specific volumes but is not given explicitly, which makes it cancel out in calculations where it appears on both sides of equations.

Understanding these properties of a cylindrical beaker helps make the problem of gas expansion more manageable.

Temperature Change

Temperature change is a fundamental concept in this problem, as it directly affects the pressure and volume of the gas within the beaker. Initially, the system is at a temperature of 273 K, which is equivalent to 0°C.
As the temperature increases, the behavior of the gas changes according to the Ideal Gas Law. This law shows that for a given amount of gas at constant pressure, volume and temperature are directly proportional. In mathematical form, it's expressed as:
\( PV = nRT \)
The increase in temperature is what causes the gas to expand and results in the mercury being pushed out of the beaker.

• Understand that a rise in temperature leads to a rise in gas volume when pressure remains constant, as per Charles's Law (\( V \propto T \).
• The challenge here is to calculate the new temperature, knowing how the gas expands with temperature and the resulting physical changes in the system.

By solving for this temperature change, it becomes clear how thermodynamics governs the behavior of gases in contained systems.

Gas Expansion

Gas expansion is the result of increasing temperature, as demonstrated by the movement of the liquid mercury within the beaker. In this situation, as half of the initial mercury amount spills out of the beaker, the volume that the gas occupies increases from 0.760 m to 1.140 m.
This expansion process abides by the laws of thermodynamics and can be quantified using the Ideal Gas Law. The formula that simplifies the relationship due to the removal of constant factors like cross-sectional area and initial pressure is:
\( \frac{V_i}{T_i} = \frac{V_f}{T_f} \)

• Gas expansion is directly linked to temperature through direct proportionality.
• As the mercury spills out, it showcases the principle that as gases expand, they need room to exert their pressure consistently, which happens by displacing the mercury and balancing external forces.

Understanding gas expansion helps to explain both the theoretical and practical ramifications of heating a gas in a confined space.

Atmospheric Pressure

Atmospheric pressure is a critical concept in this problem because it represents the external pressure exerted on both the mercury and the gas. This is especially important since the beaker is exposed to the atmosphere.
At sea level, the atmospheric pressure is standardized to 1 atmosphere (\(1\, ext{atm}\)), which is approximately equal to 101,325 Pascals. In this exercise, the atmospheric pressure keeps the system in equilibrium until heat induces changes.

• It's due to this constant pressure from the atmosphere that the Ideal Gas Law can be applied unabated without needing to adjust for pressure fluctuations.
• As pressure remains constant and only temperature changes, it simplifies the relationship between temperature and volume expansion for the gas inside the beaker.

This exercise illustrates the significance of atmospheric pressure as a controlled parameter, allowing the focus to be on changes in thermal and volumetric properties of the gas.