Review Question 15.1 Imagine that a balloon expands when brought from a cold garage into a warm room. Both the room and the garage are at atmospheric pressure. The change in the volume of the ballon is ΔV. What is the work that the air in the room does on the balloon during the process?

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Answer 1

Textbook Question

Chapter 15, Problem 1RQ

Review Question 15.1 Imagine that a balloon expands when brought from a cold garage into a warm room. Both the room and the garage are at atmospheric pressure. The change in the volume of the ballon is ΔV. What is the work that the air in the room does on the balloon during the process?

Expert Solution & Answer

To determine

The work done by air on the balloon, such that the balloon is expanding in a warm room after it is brought there from a cold garage, considering that the change in volume is ΔV and the pressure in the room as well as the garage is atmospheric pressure.

Answer to Problem 1RQ

Solution:

−PatmΔV

Explanation of Solution

Introduction:

Work done by an environment on a system under consideration has a magnitude that is evaluated by multiplying the change in volume and the constant pressure acting on the system. It is important to understand that if the volume of the system increases, the work done by the environment on the system is negative and if the volume decreases, the work done is positive.

Explanation:

Consider the provided case, in which the gas in the balloon expands. So, as the gas expands, its volume increases and the gas does positive work on the environment, that is, the air at atmospheric pressure present in the room. As a result of this, the air in the warm room does the same amount of negative work on the gas in the balloon.

The work done by the environment on the gas at constant pressure is calculated using the expression:

WEnviornment on Gas = −PatmΔV

Here, Patm is the atmospheric pressure.

Conclusion:

The work done by the air on the balloon is equal to −PatmΔV.

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Answer 2

Textbook Question

Chapter 15, Problem 1RQ

Review Question 15.1 Imagine that a balloon expands when brought from a cold garage into a warm room. Both the room and the garage are at atmospheric pressure. The change in the volume of the ballon is ΔV. What is the work that the air in the room does on the balloon during the process?

Expert Solution & Answer

To determine

The work done by air on the balloon, such that the balloon is expanding in a warm room after it is brought there from a cold garage, considering that the change in volume is ΔV and the pressure in the room as well as the garage is atmospheric pressure.

Answer to Problem 1RQ

Solution:

−PatmΔV

Explanation of Solution

Introduction:

Work done by an environment on a system under consideration has a magnitude that is evaluated by multiplying the change in volume and the constant pressure acting on the system. It is important to understand that if the volume of the system increases, the work done by the environment on the system is negative and if the volume decreases, the work done is positive.

Explanation:

Consider the provided case, in which the gas in the balloon expands. So, as the gas expands, its volume increases and the gas does positive work on the environment, that is, the air at atmospheric pressure present in the room. As a result of this, the air in the warm room does the same amount of negative work on the gas in the balloon.

The work done by the environment on the gas at constant pressure is calculated using the expression:

WEnviornment on Gas = −PatmΔV

Here, Patm is the atmospheric pressure.

Conclusion:

The work done by the air on the balloon is equal to −PatmΔV.

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Answer 3

Textbook Question

Chapter 15, Problem 1RQ

Review Question 15.1 Imagine that a balloon expands when brought from a cold garage into a warm room. Both the room and the garage are at atmospheric pressure. The change in the volume of the ballon is ΔV. What is the work that the air in the room does on the balloon during the process?

Expert Solution & Answer

To determine

The work done by air on the balloon, such that the balloon is expanding in a warm room after it is brought there from a cold garage, considering that the change in volume is ΔV and the pressure in the room as well as the garage is atmospheric pressure.

Answer to Problem 1RQ

Solution:

−PatmΔV

Explanation of Solution

Introduction:

Work done by an environment on a system under consideration has a magnitude that is evaluated by multiplying the change in volume and the constant pressure acting on the system. It is important to understand that if the volume of the system increases, the work done by the environment on the system is negative and if the volume decreases, the work done is positive.

Explanation:

Consider the provided case, in which the gas in the balloon expands. So, as the gas expands, its volume increases and the gas does positive work on the environment, that is, the air at atmospheric pressure present in the room. As a result of this, the air in the warm room does the same amount of negative work on the gas in the balloon.

The work done by the environment on the gas at constant pressure is calculated using the expression:

WEnviornment on Gas = −PatmΔV

Here, Patm is the atmospheric pressure.

Conclusion:

The work done by the air on the balloon is equal to −PatmΔV.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Textbook Question

Chapter 15, Problem 1RQ

Review Question 15.1 Imagine that a balloon expands when brought from a cold garage into a warm room. Both the room and the garage are at atmospheric pressure. The change in the volume of the ballon is ΔV. What is the work that the air in the room does on the balloon during the process?

Expert Solution & Answer

To determine

The work done by air on the balloon, such that the balloon is expanding in a warm room after it is brought there from a cold garage, considering that the change in volume is ΔV and the pressure in the room as well as the garage is atmospheric pressure.

Answer to Problem 1RQ

Solution:

−PatmΔV

Explanation of Solution

Introduction:

Work done by an environment on a system under consideration has a magnitude that is evaluated by multiplying the change in volume and the constant pressure acting on the system. It is important to understand that if the volume of the system increases, the work done by the environment on the system is negative and if the volume decreases, the work done is positive.

Explanation:

Consider the provided case, in which the gas in the balloon expands. So, as the gas expands, its volume increases and the gas does positive work on the environment, that is, the air at atmospheric pressure present in the room. As a result of this, the air in the warm room does the same amount of negative work on the gas in the balloon.

The work done by the environment on the gas at constant pressure is calculated using the expression:

WEnviornment on Gas = −PatmΔV

Here, Patm is the atmospheric pressure.

Conclusion:

The work done by the air on the balloon is equal to −PatmΔV.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

The work done by air on the balloon, such that the balloon is expanding in a warm room after it is brought there from a cold garage, considering that the change in volume is ΔV and the pressure in the room as well as the garage is atmospheric pressure.

Answer to Problem 1RQ

Solution:

−PatmΔV

Explanation of Solution

Introduction:

Work done by an environment on a system under consideration has a magnitude that is evaluated by multiplying the change in volume and the constant pressure acting on the system. It is important to understand that if the volume of the system increases, the work done by the environment on the system is negative and if the volume decreases, the work done is positive.

Explanation:

Consider the provided case, in which the gas in the balloon expands. So, as the gas expands, its volume increases and the gas does positive work on the environment, that is, the air at atmospheric pressure present in the room. As a result of this, the air in the warm room does the same amount of negative work on the gas in the balloon.

The work done by the environment on the gas at constant pressure is calculated using the expression:

WEnviornment on Gas = −PatmΔV

Here, Patm is the atmospheric pressure.

Conclusion:

The work done by the air on the balloon is equal to −PatmΔV.

Answer 8

Expert Solution & Answer

To determine

The work done by air on the balloon, such that the balloon is expanding in a warm room after it is brought there from a cold garage, considering that the change in volume is ΔV and the pressure in the room as well as the garage is atmospheric pressure.

Answer to Problem 1RQ

Solution:

−PatmΔV

Explanation of Solution

Introduction:

Work done by an environment on a system under consideration has a magnitude that is evaluated by multiplying the change in volume and the constant pressure acting on the system. It is important to understand that if the volume of the system increases, the work done by the environment on the system is negative and if the volume decreases, the work done is positive.

Explanation:

Consider the provided case, in which the gas in the balloon expands. So, as the gas expands, its volume increases and the gas does positive work on the environment, that is, the air at atmospheric pressure present in the room. As a result of this, the air in the warm room does the same amount of negative work on the gas in the balloon.

The work done by the environment on the gas at constant pressure is calculated using the expression:

WEnviornment on Gas = −PatmΔV

Here, Patm is the atmospheric pressure.

Conclusion:

The work done by the air on the balloon is equal to −PatmΔV.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

The work done by air on the balloon, such that the balloon is expanding in a warm room after it is brought there from a cold garage, considering that the change in volume is ΔV and the pressure in the room as well as the garage is atmospheric pressure.

Answer 12

Answer to Problem 1RQ

Solution:

−PatmΔV

Answer 13

Explanation of Solution

Introduction:

Work done by an environment on a system under consideration has a magnitude that is evaluated by multiplying the change in volume and the constant pressure acting on the system. It is important to understand that if the volume of the system increases, the work done by the environment on the system is negative and if the volume decreases, the work done is positive.

Explanation:

Consider the provided case, in which the gas in the balloon expands. So, as the gas expands, its volume increases and the gas does positive work on the environment, that is, the air at atmospheric pressure present in the room. As a result of this, the air in the warm room does the same amount of negative work on the gas in the balloon.

The work done by the environment on the gas at constant pressure is calculated using the expression:

WEnviornment on Gas = −PatmΔV

Here, Patm is the atmospheric pressure.

Conclusion:

The work done by the air on the balloon is equal to −PatmΔV.