**Lab Report**

*by* My Name

Student ID #

Diffusion of Gases

Teaching Assistant: TA Name

Chem 374-005

Group #6

12-6-96

Performed: 11-14-96

Lab Partner: Their Name

**Abstract**

The diffusion coefficients for CO_{2} gas and He gas were determined using a vacuum apparatus
and a modified Loschmidt tube. A rotary oil pump was used to provide the vacuum. The
composition of gases in the tube was determined by solidification of the CO_{2}, purging of the He, and
determination of the remaining CO_{2} by a pressure measurement.

__Introduction & Theory__

Diffusion is the tendency of liquids to spontaneously mix when brought into contact. In this experiment, the liquids are gasses. As molecules from one gas move into spaces where molecules from a different gas once were, the gases mix. As this mixing is based on the inherently random motion of molecules, it results in an increase in disorder and an increase of entropy. As a result, diffusion is spontaneous.

The rate at which diffusion occurs depends on an enormous variety of factors, many of which are related to the identity of the gasses taking place in the diffusion. Among these factors are forces between like and dissimilar molecules, molecular sizes, spatial considerations, etc. As a result, diffusion rates are best considered to be empirical results to be measured by experimentation for the systems in question.

**Experimental**

The procedure in *Experiments in Physical Chemistry*^{(1)} was followed in general. A few
modifications were made, generally as a matter of convenience. The copper coil was not warmed
with a water bath, but with a heat gun (used as a fan, no heat applied). Also, only one run was made
with helium only, due to time constraints.

The ambient temperature was 21.8C.

__Results__

The pressure of the CO_{2} in the top of the Loschmidt tube caused a change in the height of
mercury in the manometer, with the vacuum side at 602.0 mm and the pressurized side at 278.9 mm,
a difference of 323.1 mm Hg (323.1 Torr). Similarly, the CO_{2} pressure in the bottom of the tube was
found to be 625.3 Torr. The diffusion was allowed to take place for 35 minutes and 38 seconds.

A theoretical diffusion rate is given with the following relationship:

Where *D*_{12} is the diffusion constant, *N*_{0} is Avogadro's
number, *M*_{x} are molecular weights, and
*d*_{12} is the center to center hard sphere collision distance.

The original pressure of the CO_{2} in the system was 744.7 Torr, and 748 Torr of He. As each
gas occupied equal volumes in the tube, the pressures may be averaged to give a total pressure *p* of
746 Torr. For purposes of unit analysis, this then converts to 9.95x10^{4} N m^{-2} and finally to 9.95x10^{4}
kg m^{-1} s^{-2}.

As stated earlier, the ambient temperature *T* of the apparatus (and hence the gases) was
21.8C.

Knowing that the effective hard sphere diameters (the diameter of the general sphereical
volume the molecule occupies in space) of He, Ar, and CO_{2} are 2.58x10^{-10}, 3.42x10^{-10}, and 4.00x10^{-10}
m respectively, the radius for each can be taken as half of the given number. The center to center
distance then for any two of these molecules would be the sum of their radii. Substitution of these
values into eq. 1 yields:

The diffusion constant may be found from the experimental data with the following equation:

Where *p _{}*1 is the pressure of the CO

By neglecting the second term and solving for D, the experimental value for the diffusion constant is

| |||||

Run 1 - He | Run 2 - Ar | Run 3 - He | Run 4 - He | Run 5 - He | |

P(CO_{2}) [Pa] |
9.93x10^{4} |
1.01x10^{5} |
5.13x10^{4} |
5.15x10^{4} |
1.01x10^{5} |

P(He) [Pa] | 9.97x10^{4} |
1.01x10^{5} |
5.13x10^{4} |
5.13x10^{4} |
1.01x10^{5} |

Time (s) | 2138 | 8473 | 2130 | 2128 | 2122 |

P_{1}' (Pa) |
8.34x10^{4} |
2.39x10^{4} |
4.11x10^{4} |
1.50x10^{4} |
2.21x10^{4} |

P_{1 }(Pa) |
4.31x10^{4} |
4.68x10^{4} |
5.15x10^{4} |
2.15x10^{4} |
4.85x10^{4} |

L (m) | 0.5635 | 0.560 | 0.560 | 0.560 | 0.560 |

D (m^{2} s^{-1}) |
5.62x10^{-5} |
1.38x10^{-5} |
1.18x10^{-4} |
9.05x10^{-5} |
4.63x10^{-5} |

D (m^{2} s^{-1})
Theoretical |
4.62x10^{-5} |
1.50x10^{-5} |
8.97x10^{-5} |
8.95x10^{-5} |
4.56x10^{-5} |

Percent Error | 21.6% | 8.0% | 31.5% | 1.1% | 1.5% |

**Table 1**

__Discussion__

The most likely sources of error in this experiment are vacuum leaks, turbulence at the
stopcock, trapping of gases in the solid CO_{2}, and various approximations.

The stopcock hardly serves as an ideal method of bringing the two gases together to diffuse. As it is a mechanical device, turbulence is unavoidable, especially since the stopcock ideally should be opened and closed in a very short amount of time. Also, its design makes it a likely source of a vacuum leak, which not only introduces impurities at the location of diffusion, but additional turbulence as well.

In addition, the dimensions of the stopcock and the rest of the tube must conform to fairly small tolerances for uncalibrated glassware. Both tubes and the stopcock must be of the same diameter on the inside, and must also be the same diameter throughout their length.

A precision machined metal tube would probably serve as an improvement in the design of the apparatus used. Not only are the tolerances for metal more controllable than for glassware, but a rotatable disk with a hole in it would serve as a more ideal method of beginning and ending the diffusion process. While this would still introduce leak problems, a large, greased flange on both sides of the disk should keep this problem to a minimum. The principle advantage to a disk is it disturbs the gases very little as it is being moved, even when moved at high speeds.

In order to determine the diffusion constant of He and Ar, separation of the gases by a phase change is not very feasible due to low, similar boiling points. As both gases are inert, extraction by reaction is unlikely as well. Perhaps the most straightforward method would be to capture the gases in the top and the bottom of the tube into containers of known volume and make a density determination. While a very high precision would be required, using larger quantities of gas would help to overcome this limitation.

The concentration of the gas in each tube is related to its partial pressure is given by:

This result is equivalent to more than 180 years!

__References__

1. D. P. Shoemaker, C. W. Garland, J. W. Nibler, *Experiments in Physical Chemistry*, 6^{th}
ed., chap. 5, experiment 5, The McGraw-Hill Companies (1996).