# A Gas Law And Absolute Zero

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1 A Gas Law And Absolute Zero Equipent safety goggles, DataStudio, gas bulb with pressure gauge, 10 C to +110 C theroeter, 100 C to +50 C theroeter. Caution This experient deals with aterials that are very hot and very cold. Exercise care when handling these aterials. Do not let the get in contact with your skin or clothes. Wear safety goggles to protect your eyes fro splashes. 1 Purpose To investigate how the pressure of a given quantity of gas in a fixed volue varies with the teperature. This relationship between pressure and teperature, which is approxiately linear for low nuber densities and teperatures soewhat above liquefaction, is extrapolated to zero pressure. The teperature at which this occurs is absolute zero. 2 Description See Fig. 1. A steel bulb is filled with air at abient air pressure. The bulb is fitted with a pressure gauge that gives the absolute pressure inside the bulb in lb/in 2. When the valve is opened, the pressure gauge gives the noinal outside air pressure, about 15 lb/in 2. With the valve closed the bulb is inserted into 4 liquids of different teperatures. For each liquid, the teperature and pressure are easured. There are 5 pairs of pressure and teperature easureents. These are plotted on a graph and the best straight line is drawn. The straight line is extrapolated to zero pressure. The teperature at that point is deterined and is absolute zero. 3 Theory In an ideal gas the olecules are considered so sall that they can be approxiated as point asses. For siplicity, we let the ter olecules also include atos. It is assued that there are no forces between the olecules and that the energy of the gas is entirely the kinetic energy of the olecules. The olecules are in constant otion and as they bounce off the walls of the container exert a pressure (a force per unit area) on the walls of the container. On a very short tie scale the pressure fluctuates a great deal, but on a longer tie scale the pressure fluctuations average out. The average kinetic energy of the olecules increases with the teperature, and the higher the teperature the faster the olecules ove. The pressure exerted by the gas will increase as the teperature rises because the olecules ove faster and hit the wall harder and ore often. It is found that the pressure p and Teperature T C obey a linear relationship that can be written as p = T C + b, (1) where and b are constants and T C is the centigrade teperature. 1

2 No gas is strictly ideal and there will be deviations fro Eq.(1), particularly as the gas becoes ore dense and/or the teperature is decreased. But for dilute gases at high enough teperatures this equation is a good approxiation. As the teperature decreases ost gases will eventually becoe liquid and Eq.(1) is not valid. If data in the range where Eq.(1) is valid is extrapolated to p = 0, the very low teperature at which this occurs is independent of the gas used and is tered absolute zero. According to Eq.(1) absolute zero occurs at the centigrade teperature T C = b = There is no teperature colder than absolute zero! If the kelvin teperature T is defined as T = T C + b, Eq.(1) can be written as p = T. (2) What should the constant be, and is it a constant for all situations? Following is a rough derivation. Let the nuber density (nuber of olecules per unit volue) of the gas olecules be n N. The nuber of olecules striking per unit area of the container wall per unit tie is n N v, where v is a suitable average speed of the olecules. The oentu carried by each olecule is proportional to Mv, where M is the ass of the olecule. The oentu carried toward the wall per unit area per unit tie is proportional to n N Mv 2, and the oentu carried away fro the wall per unit area per unit tie the sae value. The net oentu transferred to the wall per unit area per unit tie will be proportional to 2n N Mv 2, and according to Newton s 2nd law, is proportional to the pressure. But Mv 2 is proportional to the kinetic energy which in turn, according to kinetic theory, is proportional to the Kelvin teperature. We ust have that the pressure is proportional to the product of the n N and T, or p = n N kt, (3) where k is the proportionality constant and is called Boltzann s constant. For Eq.(3) to be written in this siple for without an additive constant on the right hand side, the teperature ust be in kelvin. Note that one degree kelvin is equal to one degree centigrade. Let the volue of the gas be V and Avogadro s nuber be N A. If both the nuerator and denoinator of the right hand side of Eq.(3) are ultiplied by V and N A, this equation can be written P V = nrt, (4) where n is the nuber of oles of the gas and N A k = R = 8.31 J/ole/K is the gas constant. It is custoary to write K for degrees kelvin rather than K or deg K. 4 Experient Wear safety goggles to protect your eyes fro splashes. 1. Begin by depressing the pin inside the end of the valve attached to the bulb. This connects the bulb to the roo air and sets the bulb pressure at one atosphere. Do not open the valve again. Record the value indicated by the pressure gauge in units of lb/in 2, tapping the gauge gently with a fingernail to be sure that the needle is not sticking (a good procedure with any echanical gauge). Record also the teperature 2

3 in the roo using the 10 C to +110 C theroeter. Be careful that your hand does not war the bulb above roo teperature. 2. Ierse the bulb in boiling water. Use the 10 C to +110 C theroeter to easure the teperature of the water. Do not let the theroeter touch the botto or sides of the boiling water container, where the teperature ay differ fro that of the water. When equilibriu is reached (the pressure reading becoes constant), record the pressure and the teperature. Reove the bulb fro the stea bath and allow it to reach roo teperature. Check to see that the pressure reading is the sae as it was before subjecting the gas to the boiling water. If it is significantly different you ay have a leak. 3. Ierse the bulb in a water-ice ixture. Using the sae procedures as before, easure the pressure and teperature using the sae theroeter. 4. Note: In this part of the experient, do not use the sae theroeter. You will break it. Use the 100 C to +50 C theroeter. Ierse the bulb and a suitable theroeter in alcohol. At first cool the alcohol by adding sall chips of solid carbon dioxide. Do not touch the chips with unprotected fingers! As the teperature drops and the boiling becoes less violent add larger pieces of the carbon dioxide. It should be possible to lower the teperature to about 72 C. Record the pressure and teperature as before. 5. In this part of the experient, do not insert any theroeter as it will break. Ierse the bulb in liquid nitrogen. Do this slowly to iniize violent boiling. Take care not to touch the liquid nitrogen or to be splashed by it. Assue the teperature is 196 C. Allow the gas to reach equilibriu and record the pressure. 6. When the bulb returns to roo teperature check to see if it returns to about its original pressure. 5 Data Analysis Analyze your data by plotting it on a DataStudio graph display and fitting a straight line to the data points. This is a least squares fit. As the horizontal axis of the graph display will not accept negative nubers, it is necessary to rewrite Eq.(1) as T = 1 p b. (5) The teperature T will be plotted on the vertical axis and the pressure p on the horizontal axis. The slope of the line is 1 and the vertical axis intercept (when p = 0) is b, which is the value of absolute zero. To plot your data and find the slope and vertical axis intercept You re required to. 1. Open New Epty Data Table by selecting it fro Experient on the enu bar. 2. Enter your 5 pairs of data points in the x and y coluns. 3. Drag the Graph icon fro the Displays enu to the Data that you just input in the Data enu. The horizontal axis will be islabeled x but the nubers will be right. 3

4 4. Click the statistics button at the top of the graph window. Reeber to resize either axis you, double click on the graph to get Graph Settings. After click on the axis settings tab. Once here you can input the axiu and iniu values. On the horizontal axis ake ax a nuber slightly larger than your axiu recorded pressure and leave in at 0. On the vertical axis ake ax 110 and ake in All 5 of your data points should be visible in the graph window. 5. Ath the top of the graph window there is a Fit button, click it and then choose linear. The least squares straight line fit will appear through your data points. Print out the required nuber of graphs, cross out Tie(s) below the horizontal axis and enter Pressure lb/in Obtain values for and b. In doing so, give units. Consider the following questions. 6 Results What are the diensions of teperature? List the units of teperature that you know. What are the diensions of and b? What units of and b are you using? Hint: They are not the sae. Discuss your results. Suggestions. How well do your data fit a straight line? What is your value for absolute zero in degrees centigrade ( b )? Do the units ake sense? Since Boltzann s constant k = J/K, what is n N for your experient? Hint: k is given in S.I. units. Express in S.I. units, and calculate the density in olecules per cubic eter. Envision, if you can, that nuber of olecules. The diaeter of the bulb is about 10 c. Neglecting the wall thickness of the bulb, and the volue of the connecting tubing and pressure gauge, about how any oles of air are you working with? One of the assuptions in your data analysis is that the density of the air in the bulb stays constant. How any reasons can you think of for this not to be so? What kind of errors ight be introduced? Exercise. Obtain Eq.(4) fro Eq.(3). 4

5 7 Coent You ight think that at zero degrees Kelvin atoic and olecular otion ceases. This is not the case. There is still a little bit of otion to these particles at absolute zero which is called zero point energy. One of the stateents of the Heisenberg uncertainty principle of quantu echanics is that for each linear diension the product of the uncertainties in oentu and position of a particle ust be greater than or equal to Planck s constant h divided by 2π or h/2π. For exaple, if the linear diension is denoted by x and the uncertainties in position and oentu are denoted by x and p x, then x p x. If the oentu were zero (no otion) its oentu would be known exactly (no uncertainty, p x = 0) and we would have no idea where the particle was. 8 Finishing Up Please leave your bench as you found it. Thank you. Figure 1: Fixed voloue of gas in a constant teperature bath. 5

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