Kinetic Theory and the First Law of Thermodynamics

On September 4th, 2014, the Physics 4B class discussed the different applications of the Gas Laws and the properties of gases.

Work Done by and Expanding Gas:

My lab group and I thought that the equation of work was the integral of Force vector and the change of the radius vector because Force is not always constant like how we want it to be by recalling what we learned from last semester.

Below is a demonstration done by Professor Mason explaining the work concept by using a glass syringe and compressed air. It was almost impossible for Professor Mason to displace the syringe at a different volume once the syringe was clamped off from any extra air. But with enough force, he was able to keep the syringe at a certain volume marker.

My lab group and I continued to derive the work equation and it turns out that the actual work equation was the integral of pressure times the change in volume.

We were asked to describe an equation as if we were explaining it to a child. The equation was the change in energy is equal to heat subtracted by work, which is the Conservation of Energy equation. My lab group and I used the illustration of a man walking up a hill. The change in body temperature would be the heat, the physical tiredness one feels would be the change in energy, and the actual activity would be the work. Hopefully, a child would be able to understand the basic concept of the Conservation of Energy equation.

What if heat was equal to zero? What kind of action describes a situation where there is motion but not heat is given off? My lab group and I came up with the compressing of a spring and a pendulum. Both motions do not require any sort of heat to initially start the motion.

We were then asked to find the work done on a copper bar that weighed 1kg when the temperature changed from 20 degrees Celsius to 50 degrees Celsius. The pressure remained constant and the Boltzmann's Constant and the beta of the copper were given. We found the work done on the copper bar when heated to be 0.016 J.

Next, we were asked how much energy was put into the copper bar. Using the information we found in the previous question and being given the specific heat of copper, we were able to calculate the internal energy by using the Conservation of Energy equation. The answer was about 11,600 J.

Kinetic Theory:

Afterwards, we did a series of derivations. First, we found velocity total squared to be 3 times the velocity in the x-direction squared. Thus, the Force in the x-direction was found to be mass times velocity in the x-direction squared all divided by a length: X. But because pressure equals F/A, a new pressure equation was derived. Mass times velocity in the x-direction squared all divided by volume: X cubed.

If there are more than one molecule in the closed system, the pressure equation will look like this: P=N(mv^2)/3V, where N is equal to the number of molecules in the system. Below is a few clips of a closed system where molecules are stimulated because of temperature, pressure, and number of molecules. The first has only two molecules present and the second simulation has... many.

These simulations show that many molecules in a closed system means more pressure and high temperature.

Because the internal energy is equal to the Boltzmann's Constant multiplied by time T, a new ideal gas law equation is derived: PV=(2/3)NkT. If pressure is found to be constant, then the root mean squared velocity is: sqrt(3RT/M) where M is the molar mass of the molecule.

Isothermal and Adiabatic Processes for an Ideal Gas:

In an isothermal compression, the change of internal energy and the change in time are equal to zero. Thus, work equals to heat in this situation. If the following is true, then P1V1=P2V2. This means that the change in internal energy, which is equal to (3/2)NkT, is equal to zero.

In an adiabatic compression of a gas, the heat in the system is equal to zero, and thus, the change in internal energy is equal to the negative work done on the system. Earlier, we found that the change in internal energy is equal to (3/2)NkT. Then, negative work is equal to the change in internal energy just negative. Because W is equal to pressure times the change in volume, P times delta V is equal to -(3/2)NkT. If pressure is constant, then we can eliminate P and the constants N and k. We are then left with (3/2)(delta T/T)=-(delta V/V). If we integrate this equation, we get the adiabatic expansion equation for monatomic gases  that is shown in this picture:

Professor Mason's derivations:

The Fire Syringe - Fahrenheit 451:

Before we conducted the experiment, we were asked to predict if the small piece of cellulose would ignite. We measured the initial length of air column to be .991 cm, the final length of the air column to be .08 cm, the inner radius of the tube to be .0485 cm, initial volume to be 7.32 x 10^-4 cm cubed, final volume to be 5.91 x 10^-4 cm cubed, and the initial temperature to be 299 K.

We calculated the predicted final temperature to be 2055 K, which is 1781 degrees Celsius and that is hot enough for the cellulose (cotton) to ignite within the closed system.

So, we tried it with Professor Mason's guidance.

This time, one of our lab group members tried it and the cellulose did ignite on fire!

Some sources of error in this experiment was that our initial length of air column was not long enough to create a high enough pressure within the system. Thus, the initial attempt was a failed attempt.