/*! 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 19 A thermos bottle fitted with a p... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A thermos bottle fitted with a piston is filled with a gas. Since the thermos bottle is well insulated, no heat can enter or leave it. The piston is pushed in, compressing the gas. a) What happens to the pressure of the gas? Does it increase, decrease, or stay the same? b) What happens to the temperature of the gas? Does it increase, decrease, or stay the same? c) Do any other properties of the gas change?

Short Answer

Expert verified
Answer: During adiabatic compression, the pressure and temperature of the gas increase, while other properties, such as the number of moles and gas constant, remain constant.

Step by step solution

01

Recall the Ideal Gas Law

The Ideal Gas Law is given by the equation: PV = nRT where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature (in Kelvin). This equation helps us understand the relationship between the gas properties during the compression process.
02

Analyze the impact of compression on pressure

As the piston is pushed in, the volume of the gas (V) decreases. According to the Ideal Gas Law (PV = nRT), if V decreases while n, R, and T remain constant, then the pressure (P) must increase for the equation to be balanced. Therefore, the pressure of the gas increases.
03

Consider the adiabatic process

Since the thermos is well insulated, the compression occurs without any heat exchange, which is called an adiabatic process. In an adiabatic process, the following relationship holds: PV^γ = constant where γ is the adiabatic index (also known as the heat capacity ratio, equal to the ratio of specific heat capacities, Cp/Cv).
04

Analyze the impact of compression on temperature

In the adiabatic process, as the piston is pushed in and the gas volume decreases, the pressure increases. From the equation PV^γ = constant, we observe that if V decreases and P increases, T must also increase to maintain the equation's balance. Therefore, during adiabatic compression, the temperature of the gas increases.
05

Determine any other property changes

From the Ideal Gas Law and the adiabatic process equation, we can conclude that the main properties of the gas that change are pressure and temperature. Other variables, such as the number of moles (n) and the gas constant (R) remain constant during the process. So, no other properties of the gas change. In summary: a) The pressure of the gas increases. b) The temperature of the gas increases. c) No other properties of the gas change.

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!

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

Several days after the end of a snowstorm, the roof of one house is still completely covered with snow, and another house's roof has no snow cover. Which house is most likely better insulated?

In which type of process is no work done on a gas? a) isothermal b) isochoric c) isobaric d) none of the above

The thermal conductivity of fiberglass batting, which is 4.0 in thick, is \(8.0 \cdot 10^{-6} \mathrm{BTU} /\left(\mathrm{ft}^{\circ} \mathrm{F} \mathrm{s}\right) .\) What is the \(R\) value (in \(\left.\mathrm{ft}^{2}{ }^{\circ} \mathrm{F} \mathrm{h} / \mathrm{BTU}\right) ?\)

Arthur Clarke wrote an interesting short story called "A Slight Case of Sunstroke." Disgruntled football fans came to the stadium one day equipped with mirrors and were ready to barbecue the referee if he favored one team over the other. Imagine the referee to be a cylinder filled with water of mass \(60.0 \mathrm{~kg}\) at \(35.0^{\circ} \mathrm{C}\). Also imagine that this cylinder absorbs all the light reflected on it from 50,000 mirrors. If the heat capacity of water is \(4.20 \cdot 10^{3} \mathrm{~J} /\left(\mathrm{kg}^{\circ} \mathrm{C}\right),\) how long will it take to raise the temperature of the water to \(100 .{ }^{\circ} \mathrm{C}\) ? Assume that the Sun gives out \(1.00 \cdot 10^{3} \mathrm{~W} / \mathrm{m}^{2},\) the dimensions of each mirror are \(25.0 \mathrm{~cm}\) by \(25.0 \mathrm{~cm},\) and the mirrors are held at an angle of \(45.0^{\circ}\)

A \(2.0 \cdot 10^{2}\) g piece of copper at a temperature of \(450 \mathrm{~K}\) and a \(1.0 \cdot 10^{2} \mathrm{~g}\) piece of aluminum at a temperature of \(2.0 \cdot 10^{2} \mathrm{~K}\) are dropped into an insulated bucket containing \(5.0 \cdot 10^{2} \mathrm{~g}\) of water at \(280 \mathrm{~K}\). What is the equilibrium temperature of the mixture?

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.