The final volume of the gas.

Question

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Answer 1

Question

Chapter 10, Problem 6SP

(a)

To determine

The final volume of the gas.

(a)

Expert Solution

Answer to Problem 6SP

The final volume is $0.6 m3$ .

Explanation of Solution

Given info: The pressure of an ideal gas mixture is $1500 Pa$ and the temperature increased from $230 K$ to $920 K$ .

Write the equation satisfied by idea gas at two different pressure, volume and temperature

$P1V1T1=P2V2T2$ (1)

Here,

$P1$ is the initials pressure

$P2$ is the final pressure

$V1$ is the initial volume

$V2$ is the final volume

$T1$ is the initial temperature

$T2$ is the final temperature

The pressure is remains constant hence, $P1=P2$ thus, equation (1) will be rewritten as

$V1T1=V2T2$ (2)

Rearrange equation (2) to obtain an expression for final volume

$V2=V1T2T1$ (3)

Substitute $0.15 m3$ for $V1$ , $230 K$ for $T1$ and $920 K$ for $T2$ in equation (3)

$V2=0.15 m3×920 K230 K= 0.6 m3$

Conclusion:

The final volume is $0.6 m3$ .

(b)

To determine

The change in the volume for the process.

(b)

Expert Solution

Answer to Problem 6SP

The change in the volume for the process is $0.45 m3$ .

Explanation of Solution

Write the expression for the change in volume

$ΔV=V2−V1$

Substitute $0.15 m3$ for $V1$ and $0.6 m3$ for $V2$ in the above equation

$ΔV=0.6 m3−0.15 m3=0.45 m3$

Conclusion:

The change in the volume for the process is $0.45 m3$ .

(c)

To determine

The work done by the gas on the surroundings during the expansion.

(c)

Expert Solution

Answer to Problem 6SP

The work done by the gas on the surroundings during the expansion is $675 J$ .

Explanation of Solution

Given info:

Write the expression for work done in terms of volume and temperature

$W=PΔV$

Here,

$W$ is the work done

$ΔV$ is the change in volume

Substitute $1500 Pa$ for $P$ and $0.45 m3$ for $ΔV$ in the above equation

$W=1500 Pa×0.45 m3=675 J$

Conclusion:

The work done by the gas on the surroundings during the expansion is $675 J$ .

(d)

To determine

The work done if the initial volume is $0.24 m3$ and the same e temperature change is occurring.

(d)

Expert Solution

Answer to Problem 6SP

The work done will not be same if final volume will be $0.72 m3$ , if the initial volume is $0.24 m3$ .

Explanation of Solution

Given info: The initial volume is $0.24 m3$ .

Write the expression for final volume

$V2=V1T2T1$ (1)

Substitute $0.24 m3$ for $V1$ , $230 K$ for $T1$ and $920 K$ for $T2$ in the above equation

$V2=0.24 m3×920 K230 K=0.96 m3$

Write the expression for the change in volume

$ΔV=V2−V1$ (2)

Substitute $0.24 m3$ for $V1$ and $0.96 m3$ for $V2$ in the above equation

$ΔV=0.96 m3−0.24 m3=0.72 m3$

Write the expression for the work done.

$W=PΔV$

Substitute $1500 Pa$ for $P$ and $0.72 m3$ for $ΔV$ to find $W$ .

$W=1500 Pa×0.72 m3=1080 J$

Conclusion:

Therefore, the work done will be $1080 J$ and the final volume will be $0.72 m3$ , if the initial volume is $0.24 m3$ .

(e)

To determine

To explain is the same amount of gas involved in these two situations.

(e)

Expert Solution

Answer to Problem 6SP

The amount of gas involved in the two situations will be different.

Explanation of Solution

Write the expression for ideal gas equation

$PV=NkT$

Here,

$P$ is the pressure

$V$ is the volume

$N$ is the number of molecules

$k$ is the Boltzmann Constant

In this case, both pressure and temperature is remains as same, but there is a change in volume. According to the above equation, the number of molecules will be different for different values of $V$ at constant pressure and temperature.

Conclusion:

Therefore, different amount of gas will be involved in both case since, the volume is changing at constant pressure and temperature the N will also change.

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Answer 2

Question

Chapter 10, Problem 6SP

(a)

To determine

The final volume of the gas.

(a)

Expert Solution

Answer to Problem 6SP

The final volume is $0.6 m3$ .

Explanation of Solution

Given info: The pressure of an ideal gas mixture is $1500 Pa$ and the temperature increased from $230 K$ to $920 K$ .

Write the equation satisfied by idea gas at two different pressure, volume and temperature

$P1V1T1=P2V2T2$ (1)

Here,

$P1$ is the initials pressure

$P2$ is the final pressure

$V1$ is the initial volume

$V2$ is the final volume

$T1$ is the initial temperature

$T2$ is the final temperature

The pressure is remains constant hence, $P1=P2$ thus, equation (1) will be rewritten as

$V1T1=V2T2$ (2)

Rearrange equation (2) to obtain an expression for final volume

$V2=V1T2T1$ (3)

Substitute $0.15 m3$ for $V1$ , $230 K$ for $T1$ and $920 K$ for $T2$ in equation (3)

$V2=0.15 m3×920 K230 K= 0.6 m3$

Conclusion:

The final volume is $0.6 m3$ .

(b)

To determine

The change in the volume for the process.

(b)

Expert Solution

Answer to Problem 6SP

The change in the volume for the process is $0.45 m3$ .

Explanation of Solution

Write the expression for the change in volume

$ΔV=V2−V1$

Substitute $0.15 m3$ for $V1$ and $0.6 m3$ for $V2$ in the above equation

$ΔV=0.6 m3−0.15 m3=0.45 m3$

Conclusion:

The change in the volume for the process is $0.45 m3$ .

(c)

To determine

The work done by the gas on the surroundings during the expansion.

(c)

Expert Solution

Answer to Problem 6SP

The work done by the gas on the surroundings during the expansion is $675 J$ .

Explanation of Solution

Given info:

Write the expression for work done in terms of volume and temperature

$W=PΔV$

Here,

$W$ is the work done

$ΔV$ is the change in volume

Substitute $1500 Pa$ for $P$ and $0.45 m3$ for $ΔV$ in the above equation

$W=1500 Pa×0.45 m3=675 J$

Conclusion:

The work done by the gas on the surroundings during the expansion is $675 J$ .

(d)

To determine

The work done if the initial volume is $0.24 m3$ and the same e temperature change is occurring.

(d)

Expert Solution

Answer to Problem 6SP

The work done will not be same if final volume will be $0.72 m3$ , if the initial volume is $0.24 m3$ .

Explanation of Solution

Given info: The initial volume is $0.24 m3$ .

Write the expression for final volume

$V2=V1T2T1$ (1)

Substitute $0.24 m3$ for $V1$ , $230 K$ for $T1$ and $920 K$ for $T2$ in the above equation

$V2=0.24 m3×920 K230 K=0.96 m3$

Write the expression for the change in volume

$ΔV=V2−V1$ (2)

Substitute $0.24 m3$ for $V1$ and $0.96 m3$ for $V2$ in the above equation

$ΔV=0.96 m3−0.24 m3=0.72 m3$

Write the expression for the work done.

$W=PΔV$

Substitute $1500 Pa$ for $P$ and $0.72 m3$ for $ΔV$ to find $W$ .

$W=1500 Pa×0.72 m3=1080 J$

Conclusion:

Therefore, the work done will be $1080 J$ and the final volume will be $0.72 m3$ , if the initial volume is $0.24 m3$ .

(e)

To determine

To explain is the same amount of gas involved in these two situations.

(e)

Expert Solution

Answer to Problem 6SP

The amount of gas involved in the two situations will be different.

Explanation of Solution

Write the expression for ideal gas equation

$PV=NkT$

Here,

$P$ is the pressure

$V$ is the volume

$N$ is the number of molecules

$k$ is the Boltzmann Constant

In this case, both pressure and temperature is remains as same, but there is a change in volume. According to the above equation, the number of molecules will be different for different values of $V$ at constant pressure and temperature.

Conclusion:

Therefore, different amount of gas will be involved in both case since, the volume is changing at constant pressure and temperature the N will also change.

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Answer 3

Question

Chapter 10, Problem 6SP

(a)

To determine

The final volume of the gas.

(a)

Expert Solution

Answer to Problem 6SP

The final volume is $0.6 m3$ .

Explanation of Solution

Given info: The pressure of an ideal gas mixture is $1500 Pa$ and the temperature increased from $230 K$ to $920 K$ .

Write the equation satisfied by idea gas at two different pressure, volume and temperature

$P1V1T1=P2V2T2$ (1)

Here,

$P1$ is the initials pressure

$P2$ is the final pressure

$V1$ is the initial volume

$V2$ is the final volume

$T1$ is the initial temperature

$T2$ is the final temperature

The pressure is remains constant hence, $P1=P2$ thus, equation (1) will be rewritten as

$V1T1=V2T2$ (2)

Rearrange equation (2) to obtain an expression for final volume

$V2=V1T2T1$ (3)

Substitute $0.15 m3$ for $V1$ , $230 K$ for $T1$ and $920 K$ for $T2$ in equation (3)

$V2=0.15 m3×920 K230 K= 0.6 m3$

Conclusion:

The final volume is $0.6 m3$ .

(b)

To determine

The change in the volume for the process.

(b)

Expert Solution

Answer to Problem 6SP

The change in the volume for the process is $0.45 m3$ .

Explanation of Solution

Write the expression for the change in volume

$ΔV=V2−V1$

Substitute $0.15 m3$ for $V1$ and $0.6 m3$ for $V2$ in the above equation

$ΔV=0.6 m3−0.15 m3=0.45 m3$

Conclusion:

The change in the volume for the process is $0.45 m3$ .

(c)

To determine

The work done by the gas on the surroundings during the expansion.

(c)

Expert Solution

Answer to Problem 6SP

The work done by the gas on the surroundings during the expansion is $675 J$ .

Explanation of Solution

Given info:

Write the expression for work done in terms of volume and temperature

$W=PΔV$

Here,

$W$ is the work done

$ΔV$ is the change in volume

Substitute $1500 Pa$ for $P$ and $0.45 m3$ for $ΔV$ in the above equation

$W=1500 Pa×0.45 m3=675 J$

Conclusion:

The work done by the gas on the surroundings during the expansion is $675 J$ .

(d)

To determine

The work done if the initial volume is $0.24 m3$ and the same e temperature change is occurring.

(d)

Expert Solution

Answer to Problem 6SP

The work done will not be same if final volume will be $0.72 m3$ , if the initial volume is $0.24 m3$ .

Explanation of Solution

Given info: The initial volume is $0.24 m3$ .

Write the expression for final volume

$V2=V1T2T1$ (1)

Substitute $0.24 m3$ for $V1$ , $230 K$ for $T1$ and $920 K$ for $T2$ in the above equation

$V2=0.24 m3×920 K230 K=0.96 m3$

Write the expression for the change in volume

$ΔV=V2−V1$ (2)

Substitute $0.24 m3$ for $V1$ and $0.96 m3$ for $V2$ in the above equation

$ΔV=0.96 m3−0.24 m3=0.72 m3$

Write the expression for the work done.

$W=PΔV$

Substitute $1500 Pa$ for $P$ and $0.72 m3$ for $ΔV$ to find $W$ .

$W=1500 Pa×0.72 m3=1080 J$

Conclusion:

Therefore, the work done will be $1080 J$ and the final volume will be $0.72 m3$ , if the initial volume is $0.24 m3$ .

(e)

To determine

To explain is the same amount of gas involved in these two situations.

(e)

Expert Solution

Answer to Problem 6SP

The amount of gas involved in the two situations will be different.

Explanation of Solution

Write the expression for ideal gas equation

$PV=NkT$

Here,

$P$ is the pressure

$V$ is the volume

$N$ is the number of molecules

$k$ is the Boltzmann Constant

In this case, both pressure and temperature is remains as same, but there is a change in volume. According to the above equation, the number of molecules will be different for different values of $V$ at constant pressure and temperature.

Conclusion:

Therefore, different amount of gas will be involved in both case since, the volume is changing at constant pressure and temperature the N will also change.

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Answer 4

Question

Chapter 10, Problem 6SP

(a)

To determine

The final volume of the gas.

(a)

Expert Solution

Answer to Problem 6SP

The final volume is $0.6 m3$ .

Explanation of Solution

Given info: The pressure of an ideal gas mixture is $1500 Pa$ and the temperature increased from $230 K$ to $920 K$ .

Write the equation satisfied by idea gas at two different pressure, volume and temperature

$P1V1T1=P2V2T2$ (1)

Here,

$P1$ is the initials pressure

$P2$ is the final pressure

$V1$ is the initial volume

$V2$ is the final volume

$T1$ is the initial temperature

$T2$ is the final temperature

The pressure is remains constant hence, $P1=P2$ thus, equation (1) will be rewritten as

$V1T1=V2T2$ (2)

Rearrange equation (2) to obtain an expression for final volume

$V2=V1T2T1$ (3)

Substitute $0.15 m3$ for $V1$ , $230 K$ for $T1$ and $920 K$ for $T2$ in equation (3)

$V2=0.15 m3×920 K230 K= 0.6 m3$

Conclusion:

The final volume is $0.6 m3$ .

(b)

To determine

The change in the volume for the process.

(b)

Expert Solution

Answer to Problem 6SP

The change in the volume for the process is $0.45 m3$ .

Explanation of Solution

Write the expression for the change in volume

$ΔV=V2−V1$

Substitute $0.15 m3$ for $V1$ and $0.6 m3$ for $V2$ in the above equation

$ΔV=0.6 m3−0.15 m3=0.45 m3$

Conclusion:

The change in the volume for the process is $0.45 m3$ .

(c)

To determine

The work done by the gas on the surroundings during the expansion.

(c)

Expert Solution

Answer to Problem 6SP

The work done by the gas on the surroundings during the expansion is $675 J$ .

Explanation of Solution

Given info:

Write the expression for work done in terms of volume and temperature

$W=PΔV$

Here,

$W$ is the work done

$ΔV$ is the change in volume

Substitute $1500 Pa$ for $P$ and $0.45 m3$ for $ΔV$ in the above equation

$W=1500 Pa×0.45 m3=675 J$

Conclusion:

The work done by the gas on the surroundings during the expansion is $675 J$ .

(d)

To determine

The work done if the initial volume is $0.24 m3$ and the same e temperature change is occurring.

(d)

Expert Solution

Answer to Problem 6SP

The work done will not be same if final volume will be $0.72 m3$ , if the initial volume is $0.24 m3$ .

Explanation of Solution

Given info: The initial volume is $0.24 m3$ .

Write the expression for final volume

$V2=V1T2T1$ (1)

Substitute $0.24 m3$ for $V1$ , $230 K$ for $T1$ and $920 K$ for $T2$ in the above equation

$V2=0.24 m3×920 K230 K=0.96 m3$

Write the expression for the change in volume

$ΔV=V2−V1$ (2)

Substitute $0.24 m3$ for $V1$ and $0.96 m3$ for $V2$ in the above equation

$ΔV=0.96 m3−0.24 m3=0.72 m3$

Write the expression for the work done.

$W=PΔV$

Substitute $1500 Pa$ for $P$ and $0.72 m3$ for $ΔV$ to find $W$ .

$W=1500 Pa×0.72 m3=1080 J$

Conclusion:

Therefore, the work done will be $1080 J$ and the final volume will be $0.72 m3$ , if the initial volume is $0.24 m3$ .

(e)

To determine

To explain is the same amount of gas involved in these two situations.

(e)

Expert Solution

Answer to Problem 6SP

The amount of gas involved in the two situations will be different.

Explanation of Solution

Write the expression for ideal gas equation

$PV=NkT$

Here,

$P$ is the pressure

$V$ is the volume

$N$ is the number of molecules

$k$ is the Boltzmann Constant

In this case, both pressure and temperature is remains as same, but there is a change in volume. According to the above equation, the number of molecules will be different for different values of $V$ at constant pressure and temperature.

Conclusion:

Therefore, different amount of gas will be involved in both case since, the volume is changing at constant pressure and temperature the N will also change.

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

Answer 5

Chapter 10, Problem 6SP

(a)

To determine

The final volume of the gas.

(a)

Expert Solution

Answer to Problem 6SP

The final volume is $0.6 m3$ .

Explanation of Solution

Given info: The pressure of an ideal gas mixture is $1500 Pa$ and the temperature increased from $230 K$ to $920 K$ .

Write the equation satisfied by idea gas at two different pressure, volume and temperature

$P1V1T1=P2V2T2$ (1)

Here,

$P1$ is the initials pressure

$P2$ is the final pressure

$V1$ is the initial volume

$V2$ is the final volume

$T1$ is the initial temperature

$T2$ is the final temperature

The pressure is remains constant hence, $P1=P2$ thus, equation (1) will be rewritten as

$V1T1=V2T2$ (2)

Rearrange equation (2) to obtain an expression for final volume

$V2=V1T2T1$ (3)

Substitute $0.15 m3$ for $V1$ , $230 K$ for $T1$ and $920 K$ for $T2$ in equation (3)

$V2=0.15 m3×920 K230 K= 0.6 m3$

Conclusion:

The final volume is $0.6 m3$ .

(b)

To determine

The change in the volume for the process.

(b)

Expert Solution

Answer to Problem 6SP

The change in the volume for the process is $0.45 m3$ .

Explanation of Solution

Write the expression for the change in volume

$ΔV=V2−V1$

Substitute $0.15 m3$ for $V1$ and $0.6 m3$ for $V2$ in the above equation

$ΔV=0.6 m3−0.15 m3=0.45 m3$

Conclusion:

The change in the volume for the process is $0.45 m3$ .

(c)

To determine

The work done by the gas on the surroundings during the expansion.

(c)

Expert Solution

Answer to Problem 6SP

The work done by the gas on the surroundings during the expansion is $675 J$ .

Explanation of Solution

Given info:

Write the expression for work done in terms of volume and temperature

$W=PΔV$

Here,

$W$ is the work done

$ΔV$ is the change in volume

Substitute $1500 Pa$ for $P$ and $0.45 m3$ for $ΔV$ in the above equation

$W=1500 Pa×0.45 m3=675 J$

Conclusion:

The work done by the gas on the surroundings during the expansion is $675 J$ .

(d)

To determine

The work done if the initial volume is $0.24 m3$ and the same e temperature change is occurring.

(d)

Expert Solution

Answer to Problem 6SP

The work done will not be same if final volume will be $0.72 m3$ , if the initial volume is $0.24 m3$ .

Explanation of Solution

Given info: The initial volume is $0.24 m3$ .

Write the expression for final volume

$V2=V1T2T1$ (1)

Substitute $0.24 m3$ for $V1$ , $230 K$ for $T1$ and $920 K$ for $T2$ in the above equation

$V2=0.24 m3×920 K230 K=0.96 m3$

Write the expression for the change in volume

$ΔV=V2−V1$ (2)

Substitute $0.24 m3$ for $V1$ and $0.96 m3$ for $V2$ in the above equation

$ΔV=0.96 m3−0.24 m3=0.72 m3$

Write the expression for the work done.

$W=PΔV$

Substitute $1500 Pa$ for $P$ and $0.72 m3$ for $ΔV$ to find $W$ .

$W=1500 Pa×0.72 m3=1080 J$

Conclusion:

Therefore, the work done will be $1080 J$ and the final volume will be $0.72 m3$ , if the initial volume is $0.24 m3$ .

(e)

To determine

To explain is the same amount of gas involved in these two situations.

(e)

Expert Solution

Answer to Problem 6SP

The amount of gas involved in the two situations will be different.

Explanation of Solution

Write the expression for ideal gas equation

$PV=NkT$

Here,

$P$ is the pressure

$V$ is the volume

$N$ is the number of molecules

$k$ is the Boltzmann Constant

In this case, both pressure and temperature is remains as same, but there is a change in volume. According to the above equation, the number of molecules will be different for different values of $V$ at constant pressure and temperature.

Conclusion:

Therefore, different amount of gas will be involved in both case since, the volume is changing at constant pressure and temperature the N will also change.

Answer 6

Chapter 10, Problem 6SP

(a)

To determine

The final volume of the gas.

(a)

Expert Solution

Answer to Problem 6SP

The final volume is $0.6 m3$ .

Explanation of Solution

Given info: The pressure of an ideal gas mixture is $1500 Pa$ and the temperature increased from $230 K$ to $920 K$ .

Write the equation satisfied by idea gas at two different pressure, volume and temperature

$P1V1T1=P2V2T2$ (1)

Here,

$P1$ is the initials pressure

$P2$ is the final pressure

$V1$ is the initial volume

$V2$ is the final volume

$T1$ is the initial temperature

$T2$ is the final temperature

The pressure is remains constant hence, $P1=P2$ thus, equation (1) will be rewritten as

$V1T1=V2T2$ (2)

Rearrange equation (2) to obtain an expression for final volume

$V2=V1T2T1$ (3)

Substitute $0.15 m3$ for $V1$ , $230 K$ for $T1$ and $920 K$ for $T2$ in equation (3)

$V2=0.15 m3×920 K230 K= 0.6 m3$

Conclusion:

The final volume is $0.6 m3$ .

Answer 7

Chapter 10, Problem 6SP

(a)

To determine

The final volume of the gas.

Answer 8

(a)

Expert Solution

Answer to Problem 6SP

The final volume is $0.6 m3$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given info: The pressure of an ideal gas mixture is $1500 Pa$ and the temperature increased from $230 K$ to $920 K$ .

Write the equation satisfied by idea gas at two different pressure, volume and temperature

$P1V1T1=P2V2T2$ (1)

Here,

$P1$ is the initials pressure

$P2$ is the final pressure

$V1$ is the initial volume

$V2$ is the final volume

$T1$ is the initial temperature

$T2$ is the final temperature

The pressure is remains constant hence, $P1=P2$ thus, equation (1) will be rewritten as

$V1T1=V2T2$ (2)

Rearrange equation (2) to obtain an expression for final volume

$V2=V1T2T1$ (3)

Substitute $0.15 m3$ for $V1$ , $230 K$ for $T1$ and $920 K$ for $T2$ in equation (3)

$V2=0.15 m3×920 K230 K= 0.6 m3$

Conclusion:

The final volume is $0.6 m3$ .

Answer 13

(b)

To determine

The change in the volume for the process.

(b)

Expert Solution

Answer to Problem 6SP

The change in the volume for the process is $0.45 m3$ .

Explanation of Solution

Write the expression for the change in volume

$ΔV=V2−V1$

Substitute $0.15 m3$ for $V1$ and $0.6 m3$ for $V2$ in the above equation

$ΔV=0.6 m3−0.15 m3=0.45 m3$

Conclusion:

The change in the volume for the process is $0.45 m3$ .

Answer 14

(b)

To determine

The change in the volume for the process.

Answer 15

(b)

Expert Solution

Answer to Problem 6SP

The change in the volume for the process is $0.45 m3$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Write the expression for the change in volume

$ΔV=V2−V1$

Substitute $0.15 m3$ for $V1$ and $0.6 m3$ for $V2$ in the above equation

$ΔV=0.6 m3−0.15 m3=0.45 m3$

Conclusion:

The change in the volume for the process is $0.45 m3$ .

Answer 20

(c)

To determine

The work done by the gas on the surroundings during the expansion.

(c)

Expert Solution

Answer to Problem 6SP

The work done by the gas on the surroundings during the expansion is $675 J$ .

Explanation of Solution

Given info:

Write the expression for work done in terms of volume and temperature

$W=PΔV$

Here,

$W$ is the work done

$ΔV$ is the change in volume

Substitute $1500 Pa$ for $P$ and $0.45 m3$ for $ΔV$ in the above equation

$W=1500 Pa×0.45 m3=675 J$

Conclusion:

The work done by the gas on the surroundings during the expansion is $675 J$ .

Answer 21

(c)

To determine

The work done by the gas on the surroundings during the expansion.

Answer 22

(c)

Expert Solution

Answer to Problem 6SP

The work done by the gas on the surroundings during the expansion is $675 J$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given info:

Write the expression for work done in terms of volume and temperature

$W=PΔV$

Here,

$W$ is the work done

$ΔV$ is the change in volume

Substitute $1500 Pa$ for $P$ and $0.45 m3$ for $ΔV$ in the above equation

$W=1500 Pa×0.45 m3=675 J$

Conclusion:

The work done by the gas on the surroundings during the expansion is $675 J$ .

Answer 27

(d)

To determine

The work done if the initial volume is $0.24 m3$ and the same e temperature change is occurring.

(d)

Expert Solution

Answer to Problem 6SP

The work done will not be same if final volume will be $0.72 m3$ , if the initial volume is $0.24 m3$ .

Explanation of Solution

Given info: The initial volume is $0.24 m3$ .

Write the expression for final volume

$V2=V1T2T1$ (1)

Substitute $0.24 m3$ for $V1$ , $230 K$ for $T1$ and $920 K$ for $T2$ in the above equation

$V2=0.24 m3×920 K230 K=0.96 m3$

Write the expression for the change in volume

$ΔV=V2−V1$ (2)

Substitute $0.24 m3$ for $V1$ and $0.96 m3$ for $V2$ in the above equation

$ΔV=0.96 m3−0.24 m3=0.72 m3$

Write the expression for the work done.

$W=PΔV$

Substitute $1500 Pa$ for $P$ and $0.72 m3$ for $ΔV$ to find $W$ .

$W=1500 Pa×0.72 m3=1080 J$

Conclusion:

Therefore, the work done will be $1080 J$ and the final volume will be $0.72 m3$ , if the initial volume is $0.24 m3$ .

Answer 28

(d)

To determine

The work done if the initial volume is $0.24 m3$ and the same e temperature change is occurring.

Answer 29

(d)

Expert Solution

Answer to Problem 6SP

The work done will not be same if final volume will be $0.72 m3$ , if the initial volume is $0.24 m3$ .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Given info: The initial volume is $0.24 m3$ .

Write the expression for final volume

$V2=V1T2T1$ (1)

Substitute $0.24 m3$ for $V1$ , $230 K$ for $T1$ and $920 K$ for $T2$ in the above equation

$V2=0.24 m3×920 K230 K=0.96 m3$

Write the expression for the change in volume

$ΔV=V2−V1$ (2)

Substitute $0.24 m3$ for $V1$ and $0.96 m3$ for $V2$ in the above equation

$ΔV=0.96 m3−0.24 m3=0.72 m3$

Write the expression for the work done.

$W=PΔV$

Substitute $1500 Pa$ for $P$ and $0.72 m3$ for $ΔV$ to find $W$ .

$W=1500 Pa×0.72 m3=1080 J$

Conclusion:

Therefore, the work done will be $1080 J$ and the final volume will be $0.72 m3$ , if the initial volume is $0.24 m3$ .

Answer 34

(e)

To determine

To explain is the same amount of gas involved in these two situations.

(e)

Expert Solution

Answer to Problem 6SP

The amount of gas involved in the two situations will be different.

Explanation of Solution

Write the expression for ideal gas equation

$PV=NkT$

Here,

$P$ is the pressure

$V$ is the volume

$N$ is the number of molecules

$k$ is the Boltzmann Constant

In this case, both pressure and temperature is remains as same, but there is a change in volume. According to the above equation, the number of molecules will be different for different values of $V$ at constant pressure and temperature.

Conclusion:

Therefore, different amount of gas will be involved in both case since, the volume is changing at constant pressure and temperature the N will also change.

Answer 35

(e)

To determine

To explain is the same amount of gas involved in these two situations.

Answer 36

(e)

Expert Solution

Answer to Problem 6SP

The amount of gas involved in the two situations will be different.

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Write the expression for ideal gas equation

$PV=NkT$

Here,

$P$ is the pressure

$V$ is the volume

$N$ is the number of molecules

$k$ is the Boltzmann Constant

In this case, both pressure and temperature is remains as same, but there is a change in volume. According to the above equation, the number of molecules will be different for different values of $V$ at constant pressure and temperature.

Conclusion:

Therefore, different amount of gas will be involved in both case since, the volume is changing at constant pressure and temperature the N will also change.