The Physics of Everyday Phenomena
The Physics of Everyday Phenomena
8th Edition
ISBN: 9780073513904
Author: W. Thomas Griffith, Juliet Brosing Professor
Publisher: McGraw-Hill Education
Question
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Chapter 10, Problem 6SP

(a)

To determine

The final volume of the gas.

(a)

Expert Solution
Check Mark

Answer to Problem 6SP

The final volume is 0.6m3.

Explanation of Solution

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

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

V2=0.15m3×920K230K=0.6m3

Conclusion:

The final volume is 0.6m3.

(b)

To determine

The change in the volume for the process.

(b)

Expert Solution
Check Mark

Answer to Problem 6SP

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

Explanation of Solution

Write the expression for the change in volume

ΔV=V2V1

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

ΔV=0.6m30.15m3=0.45m3

Conclusion:

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

(c)

To determine

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

(c)

Expert Solution
Check Mark

Answer to Problem 6SP

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

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

W=1500Pa×0.45m3=675J

Conclusion:

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

(d)

To determine

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

(d)

Expert Solution
Check Mark

Answer to Problem 6SP

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

Explanation of Solution

Given info: The initial volume is 0.24m3.

Write the expression for final volume

V2=V1T2T1 (1)

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

V2=0.24m3×920K230K=0.96m3

Write the expression for the change in volume

ΔV=V2V1 (2)

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

ΔV=0.96m30.24m3=0.72m3

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 0.72m3, if the initial volume is 0.24m3.

(e)

To determine

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

(e)

Expert Solution
Check Mark

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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Chapter 10 Solutions

The Physics of Everyday Phenomena

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