Candle Stick
We began by taking a candle and lighting it. It was determined that when a candle is
burning that what is actually burning is the candle wax gas, not the wick! We
then took the candle and put it into a cylinder. It was predicted that the flame would get
dimmer because of the lack of oxygen, but the result was that the flame
actually went out because it burnt up all the oxygen within the cylinder. We
then placed a tube inside the cylinder and then put the candle inside the tube. It was predicted that the flame would burn
the same and the results showed that too.
This is because the tube acted like a type of chimney almost and because
the tube was sucking the CO2 from the candle smoke out of the cylinder, the
cylinder was sucking O2 back into the cylinder and down to the candle so that
the candle was able to breath. 
We then were told that a candle was going to be put into a large gallon sized glass bottle with a cork seal and then dropped six feet. We predicted that the candle would burn less brightly because it is burning up the oxygen. While the true answer was that the candle got dimmer, the reasoning was that the oxygen was being pushed to the top of the glass container and so there were no convection currents, only diffusion currents which were giving the flame a much smaller rounder shape.
ActivPhysics
We did questions on ActivPhysics to help us further
understand the isobaric process, the isochoric process, and the isothermal
process. For question 1 we were told
that for an isobaric process that the volume and temperature change and that
the number of moles and the pressure of the system stay constant. We predicted that the temperature and volume
graph would be proportional and linear because in order for the pressure to
stay constant, when the temperature increases, the volume must increase as well
because the gas molecules are bouncing around more and taking up more
space. Our prediction was correct when
checked with ActivPhysics. For question
2 we were told that for an isochoric process that the pressure and temperature
change but that the volume and the number of moles are a constant. We predicted that the pressure vs temperature
graph would be proportional and linear because since the volume is staying
constant, the increase in temperature is going to make the gas molecules move
around more and the energy they create will push against the system increase
the pressure. Our answer matched up with ActivPhysics. The next question was about isothermal
processes and it told us that the pressure and volume are changing but that the
number of moles and the temperature are constant. We predicted that the pressure vs volume
graph would be inversely proportional because if the temperature is staying the
same, then when the pressure in increased then the volume must decrease. Our
prediction matched up with ActivPhysics. 
For questions 4, 5, and 6 we were told which process was occurring and
solved problems based on what we knew about each process for example in
question 4 it told us that it is an isobaric process so we knew that pressure
and the number of moles would be constant so we used the equation V1/T1=V2/T2 and solved for the final volume which ended up being .02511 m^3 which is smaller than the initial volume because the final temperature was also smaller than the initial temperature. Because the temperature dropped, the molecules began to compress and not move so fast which in turn decreased the volume.Isothermal Process Apparatus
A gas law apparatus system was set up to display an
isothermal process. Professor Mason
slowly turned the handle on the top of the apparatus which increased the
pressure and decreased the volume of the system. The way this was an isothermal process was
because the temperature stayed at 22 degrees Celsius the entire time and it
took a long time for the whole process to be completed. After the initial and final pressures and
volumes were determined, the amount of work the professor did on the system was
found using the known facts in an isothermal process which is that the change
in internal energy is 0 so that the amount of heat added is equal to the work
done. Another known equation is that the
amount of heat added is also equal to P1V1 * ln (V2/V1) . When we were solving for the work we were
stumped because the initial pressure was 0 which would make the entire equation
0. We determined that the initial 0
pressure was the pressure of the gauge and that we were not taking into
consideration the atmospheric pressure, so we had to add the atmospheric
pressure to the initial pressure to find the true initial pressure. After calculating, we found that the work
Professor Mason did on the system was -2.58 J.
The work done is negative because it is done on a system and not by a
system. The work done is also very
small, which makes sense because Professor Mason did not have to do much work
to spin the handle around when decreasing the volume.
Heat Engine
A simple example of a heat engine could be lifting a mass
with a rubber band. When a rubber band
is heated it shrinks and in turn lift a mass that is sitting on the rubber band
upwards. We then went on to describe a
cycle that would repeatedly lift cans from a can filler conveyor belt to a can
packer conveyor belt. In four steps we
described the process as 1) the cans move forward on the conveyor belt, 2) the
cans get placed on the rubber, 3) the rubber gets heated up (Qin) to shrink the
rubber and lift the cans up and finally 4) the can gets unloaded and the rubber
gets cooled (Qout) to expand the rubber and bring it back down to its original
size so I can repeat the process. This
type of heat engine isn’t very efficient because on a hot day when the
temperature inside the building is the same temperature as the heat being added
to the rubber, then the rubber will always be contracted and will not be able
to return to its original size to pick up the cans; so on the hot day the
rubber is not able to give off heat to its surroundings (like on a normal day)
when the surroundings are as hot as the heat being added.
Tommy demonstrated another example of a heat engine which is
an isothermal process because there is a constant temperature being added to
the system (the flame). There are balls
within the tube being heated which are displacing the gas within the tube which
in turn is changing the pressure and volume of the piston attached to the gas
filled tube which is being heated. There
is not cooling system for this heat engine for the heat is constantly being
radiated from the system. The efficiency
of any heat engine is equal to the work done over the heat added to the
system. The way to make these heat engines
more efficient is to make a system extremely hot and at the same time be able
to cool it off at an extremely low temperature so that the heat added is equal
to the heat which is cooling the system which is nearly impossible and this
explains why these heat engines are so inefficient because usually the only
heat cooling the system is the heat being radiated off of the system.
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