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Consider 5.5 L of a gas at a pressure of 3.0 atm in a cylinder with a movable piston. The external pressure is changed so that the volume changes to 10.5 L. a. Calculate the work done, and indicate the correct sign. b. Use the preceding data but consider the process to occur in two steps. At the end of the first step, the volume is 7.0 L. The second step results in a final volume of 10.5 L. Calculate the work done, and indicate the correct sign. c. Calculate the work done if after the first step the volume is 8.0 L and the second step leads to a volume of 10.5 L. Does the work differ from that in part b? Explain. 7. In Question 6 the work calculated for the different conditions in the various pans of the question was different even though the system had the same initial and final conditions. Based on this information, is work a state function? a. Explain how you know that work is not a state function. b. Why does the work increase with an increase in the number of steps? c. Which two-step process resulted in more work, when the first step had the bigger change in volume or when the second step had the bigger change in volume? Explain.

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Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

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Section
BuyFindarrow_forward

Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 7, Problem 7ALQ
Textbook Problem
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Consider 5.5 L of a gas at a pressure of 3.0 atm in a cylinder with a movable piston. The external pressure is changed so that the volume changes to 10.5 L.

a. Calculate the work done, and indicate the correct sign.

b. Use the preceding data but consider the process to occur in two steps. At the end of the first step, the volume is 7.0 L. The second step results in a final volume of 10.5 L. Calculate the work done, and indicate the correct sign.

c. Calculate the work done if after the first step the volume is 8.0 L and the second step leads to a volume of 10.5 L. Does the work differ from that in part b? Explain.

7. In Question 6 the work calculated for the different conditions in the various pans of the question was different even though the system had the same initial and final conditions. Based on this information, is work a state function?

a. Explain how you know that work is not a state function.

b. Why does the work increase with an increase in the number of steps?

c. Which two-step process resulted in more work, when the first step had the bigger change in volume or when the second step had the bigger change in volume? Explain.

(a)

Interpretation Introduction

Interpretation: The work done of the given are calculated.

Concept Introduction:

Work:

  • The work is a energy transfer from one system to another to without change of entropy.
  • In gases process the work is consider as a volume change at constant pressure it is,

    W=-PΔV ......(1)PispressureΔVisvolumechange

  • The number of moles change Δn is directly proportional to volume change ΔV .
  • The number of moles change Δn or volume change ΔV is positive the work (W) is done by the system.

    The number of moles change Δn or volume change ΔV is negative the work (W) is done on the system.

Explanation of Solution

To calculate the work done in (a).

Record the given data:

          Pressure is 3.0 atm

          Initial volume in (a) step (1) =5.5L

          Final volume in (a) step (1) 10.5L

W=-P×(Finalvolume-initialvolume)W=-3atm×(10.5-5

(b)

Interpretation Introduction

Interpretation: The work done of the given are calculated.

Concept Introduction:

Work:

  • The work is a energy transfer from one system to another to without change of entropy.
  • In gases process the work is consider as a volume change at constant pressure it is,

    W=-PΔV ......(1)PispressureΔVisvolumechange

  • The number of moles change Δn is directly proportional to volume change ΔV .
  • The number of moles change Δn or volume change ΔV is positive the work (W) is done by the system.

    The number of moles change Δn or volume change ΔV is negative the work (W) is done on the system.

(c)

Interpretation Introduction

Interpretation: The work done of the given are calculated.

Concept Introduction:

Work:

  • The work is a energy transfer from one system to another to without change of entropy.
  • In gases process the work is consider as a volume change at constant pressure it is,

    W=-PΔV ......(1)PispressureΔVisvolumechange

  • The number of moles change Δn is directly proportional to volume change ΔV .
  • The number of moles change Δn or volume change ΔV is positive the work (W) is done by the system.

    The number of moles change Δn or volume change ΔV is negative the work (W) is done on the system.

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Chapter 7 Solutions

Chemistry: An Atoms First Approach
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