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Get Full Access to University Physics - 13 Edition - Chapter 19 - Problem 6e
Get Full Access to University Physics - 13 Edition - Chapter 19 - Problem 6e

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# A gas undergoes two processes. In the first, the volume

ISBN: 9780321675460 31

## Solution for problem 6E Chapter 19

University Physics | 13th Edition

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Problem 6E

A gas undergoes two processes. In the first, the volume remains constant at 0.200 m3 and the pressure increases from 2.00 X 105 Pa to 5.00 X 105 Pa. The second process is a compression to a volume of 0.120 m3 at a constant pressure of 5.00 X 105 Pa. (a) In a pV-diagram, show both processes. (b) Find the total work done by the gas during both processes.

Step-by-Step Solution:

Solution 6E Introduction We have to first draw the pv diagram for given constant volume and constant pressure process. Then we have to calculate the worked done by each process. Step 1 In the first process volume remains constant and only pressure is changing. Since the volume remains constant in this process, the line showing process will be proportional to the pressure axis. The figure is shown below. In the second process the pressure remains constant and volume is changing. Hence the graph will be parallel to the volume axis. The corresponding PV diagram is shown in the following figure.

Step 2 of 2

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