Kinetic Theory

Today we talked about work and the expansion and compression of gasses. Prof. Mason started with a set up like the one showing below. It consisted of a low-friction syringe that was connected to a rubber tube. The latter was connected to a bulb through a one-hole stopper, and the whole system was closed. There were also beakers with cool and hot water. Only one of the beakers is shown in the picture below.

The picture above was taken the moment after Prof. Mason compressed the syringe to show that it takes energy to compress a gas. This can be called doing work on the gas. He also talked about how spending energy to compress the gas ended up giving energy to the gas as it was compressed and increased the pressure and temperature of the gas. When the volume was reduced, the gas particles collided more times with the walls. This is an increase in pressure. And, with more collisions, the particles moved faster. This is an increase in temperature. Quantitatively the energy that Prof. Mason spent in compressing was gained by the gas.

Then, he submerged the bulb in cool water and let his students see the the syringe move down. The moving down of the syringe indicates a reduction of volume of the gas inside. This means that the gas was putting less pressure which means that they weren't pushing as hard against the walls of the syringe. This means that they had less energy. This was absorbed by the cool water.

Then, Prof. Mason submerged the bulb in hot water and let his students see the syringe move upwards, indicating an increase of volume. It seems like the hot water warmed the bulb and the gas inside. When the gas inside was warmed, it increased in volume. This is because the heat flowed from the water to the gas and made the gas particles move rapidly and push harder against the walls of the syringe. 

Next we calculated the change in internal energy done by a solid with a very low expanding coefficient.

First we calculated the work done by it as shown in the picture above.

Then, we calculated the heat that it gained with the formula for heat and then applied the first law of thermodynamics. This showed that some solids get to expand very little even when they gain a lot of heat. The change in internal energy is very little.

Next we explored the kinetic theory of molecular motion. In the demo, we observed the movement of two particles inside a box and saw how they collided with the walls and each other. We used this model to see how the gas law and kinetic theory think about particles.

They have the following assumptions. Every collision is completely elastic. Molecules are point like, i.e. it assumes molecules are made of only one atom. It also assumes that they are mostly empty space and have negligible mass.

When we apply these assumptions to a lot of molecules, it looks like the picture above. This means that the gas law and kinetic theory work effectively for very limited scenarios such as molecules made up of only one gas particle.

We used these assumptions to come up with formulas that we can work with that can yield measurable results.

We start with imagining an empty box of length X. A particle takes sometime to reach one wall when coming from the opposite side. We can calculate its velocity by dividing the length of the box by the time it took to travel from side to side. We do this for the x, y and z directions, and obtain the total velocity to be the square root of 3 times the velocity along the x direction. It was assumed that the velocities along every direction were equal. Then, we calculate the force on the molecule by the wall once it changes direction. Since force is equal to the change in momentum divided by the time it takes, we calculated the momentum of the particle before and after the collision and calculate the change and divide it by the time it took to do so. This yields the force to be the mass times the velocity squared divided by the length of the cube. 

Then we calculated the pressure by this particle using P = F/A and multiply it by n to expand our applicability of the formula to a gas with n moles of particles. We also transformed it to have a relationship between pressure and kinetic energy of the molecule.

Since pressure times volume can be expressed in several ways, relationships between this ways can be obtained. Our new relationship allows us to connect vrms to temperature

Next we studied a couple of processes: isothermal and adiabatic.

When a process is occurring on a gas and there is no change in temperature, all of the heat that is added or subtracted goes towards work. This yields the formulas on the bottom left of the above picture. It is also depicted that this process must be slow to allow the temperature to remain the same. Meanwhile, when a process is occurs on a gas and there is no heat that is absorbed or released, the expansion or contraction of a gas simply changes the internal energy of it. Hence we obtain the formulas on the bottom right of the above picture.

In an adiabatic process, there is no change in pressure and no heat flow interaction with the outside. Hence, the change in internal energy is equal to the work done by the gas. This equation can be reduced to the top formula on the above picture. When the expression is integrated, we obtained the bottom right expression that is true for adiabatic processes.

Next we used this formula to calculate the temperature that the inside of tube below would reach and see if it corroborated with our observations.

In the video above, we put a small piece of cotton inside a tube whose volume could be reduced quickly. Our expectation was that the quick compression of the air inside the tube would make the temperature rise and ignite the cotton. This happened wonderfully and we could observe some smoke coming out. Our calculations using the formula above showed that the final temperature inside the tube was 373.15 K. This means that cotton can be ignited at about this temperature. If instead we would've put a piece of paper inside, it wouldn't have ignited since its flash point is 451 F or about 505.93 K and this temperature wasn't reached inside the tube.

Summary
Today we talked about the work done on or by gases and saw their compression and expansion as a function of temperature. Then we calculated the change in internal energy for a solid with a very small coefficient of thermal expansion and saw that the work done by it can be negligible. Next we developed the kinetic theory and saw all of its assumptions and limitations. We developed the relationship between the root-mean-square speed and temperature of a gas, provided that they follow the ideal gas law. And finally, we developed some of the formulas for a process to be isothermal or adiabatic.

Gas Laws

We thought about what happens when an object is cooled. 

Then we filled the tube with water and marked the height of water. Then, we blew on one side and marked the new height of the other side. Then we measured the distance between the lines and the diameter of the tube.

Next we found out that the calculation of the cross-sectional area can be skipped because it cancels out.

Next we looked for the relationships of V vs T, P vs T and P vs V.

As volume increases, pressure decreases when the temperature is maintained constant. This is called Boyle's Law.

First, my group predicted that pressure and volume have an inverse relationship although we didn't know which kind. Then, we set up a syringe with 10 cc and hooked it up to the pressure sensor and computer to record the pressure as a function of volume. Then we tried to find a function that could fit this behavior. We found an equation of the form y = 1/x fits best. This means that pressure and volume have an inverse relationship. Our prediction was confirmed.

Next we tried to predict the behavior of pressure as temperature changes and maintaining temperature. Our prediction is in the following picture.

Prof. Mason run the experiment and showed us how pressure changes with temperature. It showed that pressure and temperature have a linear relationship. This is called Charles Law II.

The set up for the Volume vs Temperature. Prof. Mason brought a very special syringe that has almost no friction. Hence, it can be used to record volumes as temperature changes while maintaining the pressure.

Following is the data recorded from the set up above and shows a Temperature vs. Volume graph. It depicts that temperature and volume have a linear relationship.

When we combine all these relationships together, we get the ideal gas law. It states that pressure is inversely proportional to volume, pressure and volume are proportional to volume, and pressure times volume divided by temperature is a constant.

The triple point is a temperature and pressure that makes a substance have a solid, liquid and gas phase.
The critical point is a temperature and pressure at which a substance's physical properties vary continuously.

Next we did a diving bell problem and we figured out the pressure of air when it's submerged and the height at which the water rises to. The pressure was 2.36 * 10^6 Pa and the height was so that there was 11 cm left of air.

 Then we predicted and tested the behavior of a balloon under relatively different pressures. The first picture shows the balloon at normal pressure.

Then, we released the pressure, and the balloon expanded.

And then, we let the pressure normalize, but the balloon shrunk more than it's starting size.

The most realistic explanation for this is that the air inside the balloon escaped and that's why the balloon got smaller.

Then we tried to do the same with marshmallows.

The marshmallows started at a regular size.

Then, when the pressure decreased, the marshmallows grew in size.

But, when the pressure normalized, the marshmallows got a lot smaller than before.

Apparently, the expansion of size released the air trapped inside the marshmallows, and, when the pressure normalized, it ended up shrinking.

Next we did the high altitude balloon problem.

First a free-body diagram had to be drawn. Then, the equations could be derived.

 Working with the ideal gas law, we got an expression that could be solved for the molar mass.

We solved for the density of air at the two different heights and plugged them into the force equation.

In order to illustrate the relationships in the ideal gas law, a 3D image is needed.

Also, the ideal gas law is a lie because it only works in very restricted environments. A more exact form is the Van der Waals Equation. This is one is a second degree equation and yields more accurate measurements. There are also a third degree and a fourth degree one.

Thermal Expansion and Latent Heats

When objects heat up, most have a tendency to get larger. This is because the distance between atoms increases as they have more energy. To show this, we saw a bar that had each of its two sides made from different materials and predicted to which side would the bar bent to.

We predicted that the side that is made of invar would outgrow the side that is made out of brass and bent the bar as shown in the picture.

So Prof. Mason heated up the bar for us to see what happens.

And, he demonstrated the effect of thermal expansion on the bar. I took a video of it, but it is too big to be uploaded. Anyways, it turned out that the bar curved toward the side that is made out of invar, i.e. the brass side outgrew the invar side. He heated the bar from one side first and then cooled it to see the effect of heating the bar from the other side. The origin of the heat did not change the outcome. This shows that, even though both materials can be heated at the same time and to the same temperature, they expand at different rates.

Then Prof. Mason asked to which side would the bar curve to if it was cooled. My team's answer was:

This meant that the bar would bent towards the brass side. And, it did as Prof. Mason submitted the bar to ice for a few minutes. 

After we had seen that materials expand and contract according to their temperature, we experimented to identify a metal based on how much does it expand when heated. For this, we set up a rotary motion sensor that could measure the change in length. 

The tube was heated up using hot water vapor.

When the tube reached the temperature of the tube, the change indicated on the rotary sensor was recorded and used for calculation.

My group did the following calculation.

Then, we calculated the coefficient of thermal expansion to be 1.1 x 10^(-5) /C which turned out to be the one for steel. Hence, our tube is made of steel and we can use the coefficient to identify materials.

Next, we explored what happens when heat is added continuously to a mixture of ice and water.

And, we observed how the temperature rose as energy was added.

 We noticed at first that the beginning looks bumpy for some reason. This was unexpected. But, it seems to be due to the mixture to the energy not being distributed evenly on all the water and ice.

Then, it started rising steadily.

Until it gradually stopped rising and turned into a flat line.

By following this reaction and the amount of energy added, we can calculate how much energy it takes to melt 1 gram of ice.

So, we designed our own experiment and wrote up our plan.

We started with 99.8 g of ice and 100 g of water.

Our set up looked somewhat like professor's.

And we got the following graphs.

The first one is a graph of Temperature vs. Time.

And the second one is a graph of Heat vs. Temperature.

It took about 2 min and 28 secs. for the ice to melt. With the energy delivered being at 297.8 W, the total energy spent in melting was 44,074.4 J. From the mas of ice, we obtain the heat of fusion to be 441 J/g. Then, from looking at the slope of the second graph  we obtain the specific heat to be 4.381 J/gC. We let the water to boil for 25 secs until the heater was disconnected. And this was used to find the heat of vaporization, 2260 J/g. The percent discrepancies are 32.06 %, 4.83 % and 0% respectively.

These are the calculated uncertainties.

4.42 J/g of uncertainty belong to the heat of fusion. 659 J/g of uncertainty belong to the heat of vaporization. And, 0.07871 J/gC belong to the specific heat.

There are many possible sources of error in this experiment. Heat was also going towards the air and the cup. These quantities were not accounted for and were included as part of the behavior of water. So, this might the reason some of our values are higher than the accepted values. Also, the timing of the melting of ice and turning off the heater were probably off by a few seconds. Also, the power source was fluctuating during the experiment.