/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 27 A cylindrical glass beaker of he... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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.

Short Answer

Expert verified
The temperature is approximately 409.5 K.

Step by step solution

01

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.
02

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).
03

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.
04

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.
05

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 \( V_f = 1.140 \times A \).
06

Solve for Unknown Temperature

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

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Multiple-Concept Example 4 reviews the principles that play roles in this problem. A primitive diving bell consists of a cylindrical tank with one end open and one end closed. The tank is lowered into a freshwater lake, open end downward. Water rises into the tank, compressing the trapped air, whose temperature remains constant during the descent. The tank is brought to a halt when the distance between the surface of the water in the tank and the surface of the lake is \(40.0 \mathrm{~m}\). Atmospheric pressure at the surface of the lake is \(1.01 \times 10^{5} \mathrm{~Pa}\). Find the fraction of the tank's volume that is filled with air.

Interactive Solution \(14.55\) at provides a model for problems of this type. The temperature near the surface of the earth is \(291 \mathrm{~K}\). A xenon atom (atomic mass \(=131.29 \mathrm{u}\) ) has a kinetic energy equal to the average translational kinetic energy and is moving straight up. If the atom does not collide with any other atoms or molecules, how high up would it go before coming to rest? Assume that the acceleration due to gravity is constant throughout the ascent.

Imagine a gas diffusing through air in an arrangement like that in Figure \(14-16 a\). The diffusion rate is the number of gas atoms per second diffusing from the left to the right end of the channel. The faster the atoms move, the greater is the diffusion rate, so the diffusion rate is proportional to the rms speed of the atoms. A deuterium atom is very similar to a hydrogen atom (they both have a single electron that orbits the nucleus), except that deuterium has twice the mass. (a) If hydrogen and deuterium atoms are at the same temperature, which, if either, has the greater average kinetic energy? Give your reasoning. (b) For a given temperature, which atom, if either, would you expect to have the greater diffusion rate? Why? Problem The atomic mass of a hydrogen atom is \(1.0 \mathrm{u}\) and that of a deuterium atom is \(2.0 \mathrm{u}\). Find the ratio of the diffusion rate of hydrogen to that of deuterium. Be sure your answer is consistent with part (b) of the Concept Questions.

Suppose that a tank contains \(680 \mathrm{~m}^{3}\) of neon at an absolute pressure of \(1.01 \times 10^{5} \mathrm{~Pa}\). The temperature is changed from 293.2 to \(294.3 \mathrm{~K}\). What is the increase in the internal energy of the neon?

Hemoglobin has a molecular mass of \(64500 \mathrm{u}\). Find the mass (in \(\mathrm{kg}\) ) of one molecule of hemoglobin.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.