Please explain step by step 2) Diagrammed below is a bicycle air pump made of me
ID: 1508483 • Letter: P
Question
Please explain step by step
2) Diagrammed below is a bicycle air pump made of metal. The tube on the side is closed so no air escapes and the plunger can move up or down without letting air in or out. a) Imagine that someone in their garage pushed down the plunger and then held it in place. In terms of the equilibrium thermodynamics described by the ideal gas law (and assuming the air in the pump can be treated as an ideal gas) which quantities have changed, which have not, over the course of this process? For those quantities that have changed, describe how they have changed. Explain your reasoning. b) How would your answer to part a change (if at all) if the pump were perfectly insulated and the same process occurred? insulated and the same process occured?Explanation / Answer
A) Isothermal compression: a process performed "fast" of close to an adiabatic process since there is very little time available for the heat to flow across the system. So, when you push the plunger down quickly, it results in almost adiabatic compression initially. That decreases volume, increases pressure and temperature. But then you hold the plunger in that position and is not insulated. So heat will start flowing out (since temperature of air inside has increased after compression). This will continue till temperature equalizes to surroundings. Also pressure will decrease since according to PV=nRT, volume remaining constant, pressure has to decrease with temperature. The decrease in pressure again will upset the equilibrium and volume will decrease further till pressure inside equalised to outside (the pressure you are exerting on plunger). This will be a non quasi static process since there is a finite temperature and pressure difference across system and surrounding.
B) Adiabatic compression: if pump were insulated, the pressure and temperature will increase after compression. Now as the heat can't flow out, the system will stay as it is.
C) Isobaric compression: when pump is left alone, it will attain temperature and pressure equal to room. Then when it is put in ice bath with top exposed to surrounding, it will start losing heat to ice. So temperature will decrease till ice temperature. But as pressure is maintained constant by exposing top to surrounding, it will be an isobaric compression. Volume will decrease.