Here you can see we used the definition of velocity to find the velocity in one direction for a single molecule. Then we can use the definition of F=dp/dt=mv-(-mv)/dt=2mv/dt, We the can double the time dtx= x/Vx(the time to get to the wall) so it get back to its original location. Finally we can say Fx=mVx^2/X.
Now we can go ahead and say F=P/A, therfore P=F/A which means P=Fx/X^2=mVx^2/X^3.
Now if we want to assumer there are N molecules all we have to do is multiple by N. We can also add other directions of freedom if we let Vx^2=Vy^2=Vz^2 we can say Vtotal^2=3Vx^2. and finally we can rewrite our expression for Pressure as P=NmVtotal^2/3V, or even once more as P=2N(Ekin)/3V. (notice how we let mVtotal^2=2(Ekin))
Now from the equation up above we can say PV=2N(Ekin)/3 which we know is also equal to NkbT. We set them equal and we can find the velocity of the molecules. We can also find the a new equation by understanding the Eint can be defined as N(number of paticles) multiplies by the kinetic energy of each particle. doing this we find that Eint=3/2NkbT. On the left you see and isothermal compression where we find Q=W and on the right you see and adiabatic compression (dv=0)
Eint = -W => 3/2nKdT=-pdV we take the integral and we find the C=T^3/2V
Solving the change in internal energy, First we solve for the Work needed to raise Temperature to 50 degrees Celsius from 20 degrees Celsius of a copper bar. Then we solve for Q which is equal to mcdeltaT which we find to equal 11.58kJ and then knowing the first law of thermodynamics we get deltaE=Q-W = 11.58*10^3 J.
Here we use the equation derived above except instead of T^3/2V we use T^5/2V, this is because oxygen is diatomic and not mono-atomic. We got Tfinal = 604K or 628 degrees Fahrenheit, which we can compare to the 451 degrees needed to burn paper, which we learned in English! So we can expect the paper to burn! Don't believe me? video evidence below!
When the hot can with steam inside made contact with the room temperature water it rapidly imploded, due to the low pressure in the can relative to the pressure outside.
We experimented with a balloon and a marshmallow as to what would happen if we decreased the pressure inside the the vacuum chamber, then let back to normal.
What hypothesized above was wrong,what happened was the marshmallow much like the balloon increased in size when the pressure decreased then rapidly shrunk to a size smaller than before when the pressure was normalized.
Here is a makeshift manometer. a manometer is an instrument that uses a liquid to measure pressure!
So each side has a pressure one is known as the applied pressure and the other is our atmospheric pressure. To find the hydro static we use P=ρhg, however if we just want the pressure applied we can just subtract the reference pressure (atmospheric) pressure. We found that our applied pressure was 101026 Pa and that the mass or amount of water displaced was 5.3 * 10^-5
Here is how we think pressure and volume correspond! (which is wrong)
Here we saw Andrew pumping air into a pressure sensor!
We took measurements of the pressure sensor when at different volumes, every cubic centimeter (cc) you can see how the relationship of Pressure vs Volume is not linear but in fact are best described as inversely proportional.
Pressure Vs Temperature
This beautiful graph depicting the linear relationship between Temperature and Pressure.
If Volume is held constant you will see a linear relationship
Volume vs Temperature
Here we can see that as temperature increases the volume increases, which means they are linear.
The pressure of the air remains constant because, of the fact that Temperature and Volume are linear and as long as P=nRT/V, and volume and temperature can keep increasing pressure wont change.
As a balloon rises the gas expands!
Here we solve for mass by multiplying molar mass by the number of moles in the hot air balloon(n=PV/RT)
now we can find the volume if no helium was lost during assent by setting pv/t=PV/T=nR since the number of moles is assumed to be constant and R is a constant.
Our hypothesis which states that the brass plate will expand.
Thermal Expansion of a Invar/Brass plate.
Thermal Expansion of an Invar/Brass Plate.
A rod with a unknown coefficient of linear expansion is getting hot steam blown onto it, which then turns a pulley of known radius. This equation uses the linear thermal expansion formula however since we only have the amount of radians the pulley turned and the initial length of the rod it is slightly modified. From our solution it would be highly probable that the rod is made of steel.
A mixture of water and ice both at 0°c are heated at a constant rate. This graph shows how the latent heat of fusion ice is not so clearly defined, which means the ice doesn't all melt then start heating up but in fact the water heats up while the ice is melting.
Practice problem where we solve for the amount of water needed so that the final temperature is 22°c
Temperature vs Time Graph
In our timing we found that the ice melted in 2 mins and 28 seconds, and from there we found the latent heat of fusion of water to be 440 J/g.
Next we found let the water boil for 25 seconds. In that time we found that 3.35 g evaporated and that 7440 J were transferred into the water and that the latent heat of vaporization was 2210 J/g.
Heat/Mass vs Temperature
This graphs slope tells us the specific heat of water for our experiment, which is 4.194 [J/g-K] compared to the accepted 4.179 [J/g-K]
We used the formula for latent heat of fusion and latent heat of vaporization, then did some fancy propagation of uncertainty. However to do this we needed the uncertainty of all the components in the formula luckily i knew this from countless 4B labs and even wrote down what uncertainties we needed on the 5 steps to do this lab. (uncertainty for power was 2J to acount for the fluctuation of the machine and so to accommodate we increased it.) (Uncertainty of mass was made larger because of how the graduated cylinder is marked and because the ice was melting while we weighed it.
For this uncertainty my group and I did something clever, since we did not use the equation to find specific heat we decided it would be best to approach this by somehow incorporating our graph. What we did was jot down 8 coordinates on opposite sides of the graph from there we found the slope of the two points using y-y1/x-x1=m then we used the a standard deviation to find the uncertainty in the slope, which we defined to be the specific heat of water .
A systematic error occurs in the latent of fusion, because our equation only takes into account the ice turning into water and not the energy going into that water heating up.
A Linear graph describing the relationship between Celsius and Fahrenheit.
A methane bubble being torched!
Our hypothesis as to what would happen to the methane bubble, which was partly correct we did assume the bubble would burst into flames however we did not expect the momentum of the bubble to lift the flames upward.
A prediction and calculation where we solved for the temperature of water.
A prediction and calculation for the temperature of the water but now the mass of the objects are different.
Top: A Graph of Heat Transferring between a small liquid container holding boiling water and a large container holding room temperature water.
Bottom:Drawings demonstrating the process of heat transfer
If i were an omnipotent being...
A graph demonstrating a mixture of two liquids at different temperatures. (aka water and water)
Graphs of water being heated for roughly 20 seconds, by a Immersion heater which we measure to have a power output of 297 watts/s.
Top: Temperature vs Heat
Middle: Temperature vs Time
Bottom: Heat vs Temperature (Linear)
A graph of Heat per Unit mass vs Temperature,
The slope of this graph is describing the specific heat capacity of water, Our Experiment read this to be 4.221 J/(kg⋅oC) compared to that of the 4.179 J/(kg⋅oC) which is the scientifically accepted value.
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