Air in a bicycle pump is compressed by pushing in the handle. If the inner diameter of the pump is 3.0 cm and the pump is depressed 20. cm with a pressure of 2.00 atm,
(a) how much work is done in the compression?
(b) Is the work positive or negative with respect to the air in the pump?
(c) What is the change in internal energy of the system?
May someone explain this step by step?
Thank you!
Chapter 8 #3
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Ivy Kwok 2I
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Re: Chapter 8 #3
Before attempting the problem, it will be helpful to first convert all the given quantities into standard units! As such:
• Diameter of pump (d) = 3 cm = 0.03 m. => Radius (r) = 0.015 m.
• Distance compressed (x) = 20 cm = 0.2 m.
• Pressure applied (P) = 2 atm = 2 x 1.0 x 10^5 Pa = 2.0 x 10^5 Pa.
a) The relationship between work and distance is represented by the equation: Work = Opposing Force x Distance Moved. We are not given force, so we must calculate it using the quantities we are given. One equation we can use to represent force is P = F/A, where A is the cross-sectional area of the pump. Because the cross section is a circle, we can calculate the force by substituting our known quantities into the equation:
• P = F/A => F = PA.
• F = P(pi x r^2).
Then, we can substitute this new expression back into the original equation for work:
• W = Fx.
• W = P(pi x r^2)(x).
• W = (2.0 x 10^5 Pa) x (pi x 0.015^2 m^2) x (0.2 m).
• W = 28.274333388 J => W = 28 J (rounded to two significant figures from the given).
b) The problem asks us to analyze the work with respect to the air pump. Because the compression is being done ON the air pump, work is done ONTO it, and thus the change in work is POSITIVE.
c) Whenever work is done ONTO a system, the internal energy of the system is INCREASED. As a result, in this case, the internal energy of the system is INCREASED by 28 J.
• Diameter of pump (d) = 3 cm = 0.03 m. => Radius (r) = 0.015 m.
• Distance compressed (x) = 20 cm = 0.2 m.
• Pressure applied (P) = 2 atm = 2 x 1.0 x 10^5 Pa = 2.0 x 10^5 Pa.
a) The relationship between work and distance is represented by the equation: Work = Opposing Force x Distance Moved. We are not given force, so we must calculate it using the quantities we are given. One equation we can use to represent force is P = F/A, where A is the cross-sectional area of the pump. Because the cross section is a circle, we can calculate the force by substituting our known quantities into the equation:
• P = F/A => F = PA.
• F = P(pi x r^2).
Then, we can substitute this new expression back into the original equation for work:
• W = Fx.
• W = P(pi x r^2)(x).
• W = (2.0 x 10^5 Pa) x (pi x 0.015^2 m^2) x (0.2 m).
• W = 28.274333388 J => W = 28 J (rounded to two significant figures from the given).
b) The problem asks us to analyze the work with respect to the air pump. Because the compression is being done ON the air pump, work is done ONTO it, and thus the change in work is POSITIVE.
c) Whenever work is done ONTO a system, the internal energy of the system is INCREASED. As a result, in this case, the internal energy of the system is INCREASED by 28 J.
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