The affect on the density of the gas should be determined when the temperature of gas in a sealed, rigid container increases. Whether the results will be the same when the same experiment in a container with a movable piston at a constant external pressure takes place should be explained. Concept introduction: The results can be explained using the ideal gas law . The ideal gas equation: PV = nRT Where, V is the volume of the gas. P is the pressure of the gas. n is the number of moles of the gas. R is the gas constant. and T is the temperature.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 13, Problem 1ALQ

Interpretation Introduction

Interpretation:

The affect on the density of the gas should be determined when the temperature of gas in a sealed, rigid container increases.

Whether the results will be the same when the same experiment in a container with a movable piston at a constant external pressure takes place should be explained.

Concept introduction:

The results can be explained using the ideal gas law.

The ideal gas equation:

PV = nRT

Where, V is the volume of the gas.

P is the pressure of the gas.

n is the number of moles of the gas.

R is the gas constant.

and T is the temperature.

Expert Solution & Answer

Answer to Problem 1ALQ

In first experiment density of the gas will remain the same

In the second experiment density of the gas will decrease.

Explanation of Solution

The ideal gas equation

PV = nRT

or n/V = P/RT

Or, P/RT = d (where d is the density of the gas)

An increase in temperature will normally cause an increase in the volume. However, because the gas is enclosed in a rigid container the volume of the container cannot increase. Therefore, due to the increase in temperature, molecular momentum of air molecules also increases which further results in increase in the pressure inside container. Now, the density of a gas is the ratio of mass to its volume. If neither the mass nor the volume change as the can is heated, there will be no change in the density of the gas.

When the gas is heated in a container with a movable piston, the volume of the gas will also increase. Now, density is defined as d = m/V. Therefore, an increase in the volume will cause a decrease in the density of the gas.

Conclusion

Therefore, based on the ideal gas equation, when the gas is heated in a sealed container the density of the gas will remain constant.

However, when the gas is heated in a container with a movable piston, the density of the gas will decrease.

Want to see more full solutions like this?

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

Students have asked these similar questions

Show mechanism..don't give Ai generated solution

Don't used Ai solution

Show work. Don't give Ai generated solution

Answer 2

Question

Chapter 13, Problem 1ALQ

Interpretation Introduction

Interpretation:

The affect on the density of the gas should be determined when the temperature of gas in a sealed, rigid container increases.

Whether the results will be the same when the same experiment in a container with a movable piston at a constant external pressure takes place should be explained.

Concept introduction:

The results can be explained using the ideal gas law.

The ideal gas equation:

PV = nRT

Where, V is the volume of the gas.

P is the pressure of the gas.

n is the number of moles of the gas.

R is the gas constant.

and T is the temperature.

Expert Solution & Answer

Answer to Problem 1ALQ

In first experiment density of the gas will remain the same

In the second experiment density of the gas will decrease.

Explanation of Solution

The ideal gas equation

PV = nRT

or n/V = P/RT

Or, P/RT = d (where d is the density of the gas)

An increase in temperature will normally cause an increase in the volume. However, because the gas is enclosed in a rigid container the volume of the container cannot increase. Therefore, due to the increase in temperature, molecular momentum of air molecules also increases which further results in increase in the pressure inside container. Now, the density of a gas is the ratio of mass to its volume. If neither the mass nor the volume change as the can is heated, there will be no change in the density of the gas.

When the gas is heated in a container with a movable piston, the volume of the gas will also increase. Now, density is defined as d = m/V. Therefore, an increase in the volume will cause a decrease in the density of the gas.

Conclusion

Therefore, based on the ideal gas equation, when the gas is heated in a sealed container the density of the gas will remain constant.

However, when the gas is heated in a container with a movable piston, the density of the gas will decrease.

Want to see more full solutions like this?

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

Answer 3

Question

Chapter 13, Problem 1ALQ

Interpretation Introduction

Interpretation:

The affect on the density of the gas should be determined when the temperature of gas in a sealed, rigid container increases.

Whether the results will be the same when the same experiment in a container with a movable piston at a constant external pressure takes place should be explained.

Concept introduction:

The results can be explained using the ideal gas law.

The ideal gas equation:

PV = nRT

Where, V is the volume of the gas.

P is the pressure of the gas.

n is the number of moles of the gas.

R is the gas constant.

and T is the temperature.

Expert Solution & Answer

Answer to Problem 1ALQ

In first experiment density of the gas will remain the same

In the second experiment density of the gas will decrease.

Explanation of Solution

The ideal gas equation

PV = nRT

or n/V = P/RT

Or, P/RT = d (where d is the density of the gas)

An increase in temperature will normally cause an increase in the volume. However, because the gas is enclosed in a rigid container the volume of the container cannot increase. Therefore, due to the increase in temperature, molecular momentum of air molecules also increases which further results in increase in the pressure inside container. Now, the density of a gas is the ratio of mass to its volume. If neither the mass nor the volume change as the can is heated, there will be no change in the density of the gas.

When the gas is heated in a container with a movable piston, the volume of the gas will also increase. Now, density is defined as d = m/V. Therefore, an increase in the volume will cause a decrease in the density of the gas.

Conclusion

Therefore, based on the ideal gas equation, when the gas is heated in a sealed container the density of the gas will remain constant.

However, when the gas is heated in a container with a movable piston, the density of the gas will decrease.

Want to see more full solutions like this?

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

Answer 4

Question

Chapter 13, Problem 1ALQ

Interpretation Introduction

Interpretation:

The affect on the density of the gas should be determined when the temperature of gas in a sealed, rigid container increases.

Whether the results will be the same when the same experiment in a container with a movable piston at a constant external pressure takes place should be explained.

Concept introduction:

The results can be explained using the ideal gas law.

The ideal gas equation:

PV = nRT

Where, V is the volume of the gas.

P is the pressure of the gas.

n is the number of moles of the gas.

R is the gas constant.

and T is the temperature.

Expert Solution & Answer

Answer to Problem 1ALQ

In first experiment density of the gas will remain the same

In the second experiment density of the gas will decrease.

Explanation of Solution

The ideal gas equation

PV = nRT

or n/V = P/RT

Or, P/RT = d (where d is the density of the gas)

An increase in temperature will normally cause an increase in the volume. However, because the gas is enclosed in a rigid container the volume of the container cannot increase. Therefore, due to the increase in temperature, molecular momentum of air molecules also increases which further results in increase in the pressure inside container. Now, the density of a gas is the ratio of mass to its volume. If neither the mass nor the volume change as the can is heated, there will be no change in the density of the gas.

When the gas is heated in a container with a movable piston, the volume of the gas will also increase. Now, density is defined as d = m/V. Therefore, an increase in the volume will cause a decrease in the density of the gas.

Conclusion

Therefore, based on the ideal gas equation, when the gas is heated in a sealed container the density of the gas will remain constant.

However, when the gas is heated in a container with a movable piston, the density of the gas will decrease.

Want to see more full solutions like this?

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

Answer 5

Chapter 13, Problem 1ALQ

Interpretation Introduction

Interpretation:

The affect on the density of the gas should be determined when the temperature of gas in a sealed, rigid container increases.

Whether the results will be the same when the same experiment in a container with a movable piston at a constant external pressure takes place should be explained.

Concept introduction:

The results can be explained using the ideal gas law.

The ideal gas equation:

PV = nRT

Where, V is the volume of the gas.

P is the pressure of the gas.

n is the number of moles of the gas.

R is the gas constant.

and T is the temperature.

Expert Solution & Answer

Answer to Problem 1ALQ

In first experiment density of the gas will remain the same

In the second experiment density of the gas will decrease.

Explanation of Solution

The ideal gas equation

PV = nRT

or n/V = P/RT

Or, P/RT = d (where d is the density of the gas)

An increase in temperature will normally cause an increase in the volume. However, because the gas is enclosed in a rigid container the volume of the container cannot increase. Therefore, due to the increase in temperature, molecular momentum of air molecules also increases which further results in increase in the pressure inside container. Now, the density of a gas is the ratio of mass to its volume. If neither the mass nor the volume change as the can is heated, there will be no change in the density of the gas.

When the gas is heated in a container with a movable piston, the volume of the gas will also increase. Now, density is defined as d = m/V. Therefore, an increase in the volume will cause a decrease in the density of the gas.

Conclusion

Therefore, based on the ideal gas equation, when the gas is heated in a sealed container the density of the gas will remain constant.

However, when the gas is heated in a container with a movable piston, the density of the gas will decrease.

Answer 6

Chapter 13, Problem 1ALQ

Interpretation Introduction

Interpretation:

The affect on the density of the gas should be determined when the temperature of gas in a sealed, rigid container increases.

Whether the results will be the same when the same experiment in a container with a movable piston at a constant external pressure takes place should be explained.

Concept introduction:

The results can be explained using the ideal gas law.

The ideal gas equation:

PV = nRT

Where, V is the volume of the gas.

P is the pressure of the gas.

n is the number of moles of the gas.

R is the gas constant.

and T is the temperature.

Expert Solution & Answer

Answer to Problem 1ALQ

In first experiment density of the gas will remain the same

In the second experiment density of the gas will decrease.

Explanation of Solution

The ideal gas equation

PV = nRT

or n/V = P/RT

Or, P/RT = d (where d is the density of the gas)

An increase in temperature will normally cause an increase in the volume. However, because the gas is enclosed in a rigid container the volume of the container cannot increase. Therefore, due to the increase in temperature, molecular momentum of air molecules also increases which further results in increase in the pressure inside container. Now, the density of a gas is the ratio of mass to its volume. If neither the mass nor the volume change as the can is heated, there will be no change in the density of the gas.

When the gas is heated in a container with a movable piston, the volume of the gas will also increase. Now, density is defined as d = m/V. Therefore, an increase in the volume will cause a decrease in the density of the gas.

Conclusion

Therefore, based on the ideal gas equation, when the gas is heated in a sealed container the density of the gas will remain constant.

However, when the gas is heated in a container with a movable piston, the density of the gas will decrease.

Answer 7

Chapter 13, Problem 1ALQ

Interpretation Introduction

Interpretation:

The affect on the density of the gas should be determined when the temperature of gas in a sealed, rigid container increases.

Whether the results will be the same when the same experiment in a container with a movable piston at a constant external pressure takes place should be explained.

Concept introduction:

The results can be explained using the ideal gas law.

The ideal gas equation:

PV = nRT

Where, V is the volume of the gas.

P is the pressure of the gas.

n is the number of moles of the gas.

R is the gas constant.

and T is the temperature.

Answer 8

Expert Solution & Answer

Answer to Problem 1ALQ

In first experiment density of the gas will remain the same

In the second experiment density of the gas will decrease.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

The ideal gas equation

PV = nRT

or n/V = P/RT

Or, P/RT = d (where d is the density of the gas)

An increase in temperature will normally cause an increase in the volume. However, because the gas is enclosed in a rigid container the volume of the container cannot increase. Therefore, due to the increase in temperature, molecular momentum of air molecules also increases which further results in increase in the pressure inside container. Now, the density of a gas is the ratio of mass to its volume. If neither the mass nor the volume change as the can is heated, there will be no change in the density of the gas.

When the gas is heated in a container with a movable piston, the volume of the gas will also increase. Now, density is defined as d = m/V. Therefore, an increase in the volume will cause a decrease in the density of the gas.