Physics for Scientists and Engineers: Foundations and Connections
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN: 9781133939146
Author: Katz, Debora M.
Publisher: Cengage Learning
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GOAL Calculate work from a PV diagram.
PROBLEM Find the numeric value of the work
done on the gas in (a) Figure a, and (b)
Figure b
STRATEGY The regions in question are
composed of rectangles and triangles. Use basic
geometric formulas to find the area underneath
each curve. Check the direction of the arrow to
determine signs.
10P)
300
200-
5.00-
200
P
₤4. Gd
300
100 2.00 1.00
40
PPP
2005
-Vin
100
3,00
100 200 500
PIP)
200-
倒
-Vou
1.00
100
-Vi
1.00 2.00 1.00
(c)
100 200 500
-Vim³
(4)
SOLUTION
(A) Find the work done on the gas in Figure a.
Compute the areas 41 and 42 in
Figure a. 4, is a rectangle and 42 is a
triangle.
Sum the areas (the arrows point to
Increasing volume, so the work done
on the gas is negative).
(B) Find the work done on the gas in Figure b.
Compute the areas of the two
rectangular regions.
Sum the areas (the arrows point to
decreasing volume, so the work done
on the gas is positive).
LEARN MORE
41-height x width
41 (1.00 x 10 Pa)(2.00 m³) -2.00 x 105
42-base x height
42-(2.00 m³ (2.00 x 105 Pa) -2.00 x 105]
Area 41 +42-4.00 x 10
--4.00 x 105
41-height x width
4-(1.00 x 10 Pa)(1.00 m³) - 1.00 × 105
4,-height x width
42-(2.00 x 105 Pa)(1.00 m³) - 2.00 x 105
Area 41 +42-3.00 x 105
- +3.00×1051
REMARKS Notice that in both cases the paths in the DV diagrams start and end at the same points, but
the answers are different.
QUESTION IS work done on a system during a process in which its volume remains constant? Explain.
(Select all that apply.)
Changing the pressure of a gas at constant volume does work.
Heating a gas at constant volume does work.
Work requires not only the force from pressure acting over an area, but also a displacement.
No work is done during such a process.
Work is done during such a process.
EXERCISE
HINTS:
GETTING STARTED | IM STUCK!
Compute the work done on the system in Figure c from -1.50 m³ to - 3 m³.
Compute the work done on the system in Figure d from -2.50 m³ to -1 m³.
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Transcribed Image Text:GOAL Calculate work from a PV diagram. PROBLEM Find the numeric value of the work done on the gas in (a) Figure a, and (b) Figure b STRATEGY The regions in question are composed of rectangles and triangles. Use basic geometric formulas to find the area underneath each curve. Check the direction of the arrow to determine signs. 10P) 300 200- 5.00- 200 P ₤4. Gd 300 100 2.00 1.00 40 PPP 2005 -Vin 100 3,00 100 200 500 PIP) 200- 倒 -Vou 1.00 100 -Vi 1.00 2.00 1.00 (c) 100 200 500 -Vim³ (4) SOLUTION (A) Find the work done on the gas in Figure a. Compute the areas 41 and 42 in Figure a. 4, is a rectangle and 42 is a triangle. Sum the areas (the arrows point to Increasing volume, so the work done on the gas is negative). (B) Find the work done on the gas in Figure b. Compute the areas of the two rectangular regions. Sum the areas (the arrows point to decreasing volume, so the work done on the gas is positive). LEARN MORE 41-height x width 41 (1.00 x 10 Pa)(2.00 m³) -2.00 x 105 42-base x height 42-(2.00 m³ (2.00 x 105 Pa) -2.00 x 105] Area 41 +42-4.00 x 10 --4.00 x 105 41-height x width 4-(1.00 x 10 Pa)(1.00 m³) - 1.00 × 105 4,-height x width 42-(2.00 x 105 Pa)(1.00 m³) - 2.00 x 105 Area 41 +42-3.00 x 105 - +3.00×1051 REMARKS Notice that in both cases the paths in the DV diagrams start and end at the same points, but the answers are different. QUESTION IS work done on a system during a process in which its volume remains constant? Explain. (Select all that apply.) Changing the pressure of a gas at constant volume does work. Heating a gas at constant volume does work. Work requires not only the force from pressure acting over an area, but also a displacement. No work is done during such a process. Work is done during such a process. EXERCISE HINTS: GETTING STARTED | IM STUCK! Compute the work done on the system in Figure c from -1.50 m³ to - 3 m³. Compute the work done on the system in Figure d from -2.50 m³ to -1 m³.
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