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Research Question and Hypothesis
Research Question and Hypothesis
How does the type gas in a latex balloon affect the rate at which gasses effuse through the balloon? My prediction is that the larger the molecule, the longer it will take to effuse.
Standards
Standards
HS-PS1-3 Matter and its Interactions
PS1A: Structure and Properties of Matter
PS3D: Energy in Chemical Processes and Everyday Life
Experimental Design
Experimental Design
Three latex balloons will be filled with three different gasses; Helium, Hydrogen, and Carbon Dioxide. The balloons will be monitored and measured throughout the day. The balloons are not spheres but are close to ellipsoids, so the volume of the balloon can be estimated by measuring the Height/2(a), Length/2(b), and Width/2(c).
Independent variable
Independent variable
The independent variable is time and will be measured using a clock.
Dependent variables
Dependent variables
The dependent variable is volume of the balloon.
Series
Series
Study the change in volume of a balloon as a function of time with various gases and balloon material.
Constants and Controls
Constants and Controls
The balloons will be identical, expect for the mylar, and will be stored in the same fume hood.
Materials
Materials
Latex balloons can be bought at a party supply store. Helium can also be purchased at a supply store. If you don't have a supply of hydrogen and carbon dioxide gas, the hydrogen can be created by dropping zinc into a flask of hydrochloric acid and putting the balloon over the top. Similarly carbon dioxide can be created with baking soda and vinegar in a flask.
Meter sticks to measure the dimensions.
My students helped me create a better tool for measuring the dimensions of the balloon. We attached a right angle ruler to the meter stick so we could press the balloon up to the edge.
Procedures
Procedures
Fill a latex balloon with helium.
Fill a latex balloon with hydrogen gas.
Fill a latex balloon with carbon dioxide gas.
The balloons should be about the same size. You can use a fabric ruler to measure the balloon size.
Tape all four balloons to the inside of a fume hood.
Each hour measure the length, width, and height of the balloons
Sample data and graphs
Sample data and graphs
Analysis & Conclusions
While recording data, I noticed that after a while the balloons stopped shrinking. The data confirmed this observation. It looked like there was an inverse relationship between the volume and the time. This means the the rate of effusion decreases as the volume of the balloon shrinks. I tried graphing the inverse of the volume vs the time to see if I could linearize the data, but the data still didn't look great.
The balloons never completely deflated, so to dispose of the hydrogen filled balloons, I ignited them over the a flame. However to my surprise the balloon didn't pop.
Here is a slow motion video of the first balloon. No explosion.
I tried the second hydrogen balloon. A small hole burned into the balloon and the gas started spewing out. As you can see in the video, it didn't ignite. The gas wasn't hydrogen.
So it occurred to me that as the hydrogen gas was effusing out of the balloon, air must have been effusing back in. This would explain why the rate of effusion seemed to slow down near the end of the experiment. At some point most of the helium must have been displaced by the surrounding air, and the rate of air leaving was just about the same as the air entering. I used a different graphing program to quickly be able to graph the data and find the rate of effusion for just the first group of measurements.
Hydrogen trial 1 Hydrogen trial 2
Does the data support Graham's Law of Effusion?
According to Graham's Law of Effusion, the ratio of the rates of effusion should be equal to the square root of the inverse of their molar masses.
The following is a sample calculation of the Helium and the Hydrogen (trial2)
The experimental rate ratio is really close to the square root the masses!
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