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I was doing an experiment to inflate a bag using CO2 produced from a reaction between baking soda and vinegar. I came up with a balanced chemical equation and used the ideal gas law to calculate the amount of baking soda and vinegar I would need to fully inflate the bag. This worked great, but now I need to do something else: find the pressure inside the bag. Here's my process:

  1. Weigh the bag with gas inside it, then deflate it and weigh it again. The difference in weight is the grams of CO2 produced as gas.

  2. This yielded 0.66g. Using the molar mass of CO2, I calculated roughly 0.014 moles of CO2 gas produced. I plugged this into the ideal gas law equation along with gas constant 0.0821, temperature of 295K, and volume 0.75L for my bag.

  3. Solving this yields a pressure of 0.48 atm. That's impossible, since the bag cannot inflate unless it has a pressure greater than the pressure of its environment (around 1.0 ATM).

What did I do wrong here?

(Disclosure - also asked at https://www.reddit.com/r/chemhelp/comments/1bkmm0j/calculating_pressure_inside_a_bag_using_the_ideal/ )

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    $\begingroup$ Did you calculate with the bag buoyancy? The difference of the weights is not the mass of CO2. $\endgroup$
    – Poutnik
    Mar 22 at 4:10
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    $\begingroup$ Note that e.g 440 mg CO2 at atmospheric pressure and temperature has weight(= what scales show, i. e. not mass) not 440 g but about 440 g - 288 g = 152 g. $\endgroup$
    – Poutnik
    Mar 22 at 14:53

2 Answers 2

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Unless that thin bag burst, the pressure was approximately one atmosphere (depending on the elevation and weather at the time), because the bag would be too weak to maintain a difference between inside and outside pressure. Therefore, consider that the volume of $\ce{CO2}$ collected was less than the theoretical capacity of the bag, 0.75 L. I.E., the bag was not fully inflated.

Also, as Poutnik mentions, consider that $\ce{CO2}$ in the bag displaces air, a mix of $\ce{N2}$, $\ce{O2}$, $\ce{Ar}$, variable amounts of $\ce{H2O}$, and other trace gases -- about 1.2 kg/m3. Subtract that from the absolute density of $\ce{CO2}$ to account for buoancy.

Do you have time and facilities to perform a further experiment? You could get an idea of the actual volume of gas generated by measuring it by displacement of water.

Gas collection by water displacement

After repeating the original experiment, carefully tie the inflated bag around a rubber tube, the end of which is immersed in water, underneath a measuring container, also filled with water, as shown above. Squeeze the bag so that the gas inside goes through the tube, bubbling up inside the inverted graduated container. Though you might not have a one liter graduated cylinder, a large, clear measuring cup, or even a plastic bottle that has been roughly calibrated, could be used to measure the $\ce{CO2}$.

Ask your instructor about performing this -- personally, I'd be pleased to have a student enhance an experiment.

Don't expect great accuracy -- some $\ce{CO2}$ dissolves in water, some will be lost transferring and tying the bag, some will remain in the bag and tube -- but it will give an idea of the volume, which should agree better with your calculation.

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Here is a thought experiment. Take a bag, fill it with air, close it and measure the mass. Then, open the bag a bit, and measure it again. Then turn it around so that the opening is on the bottom, and measure again. Finally, squeeze the air out of the bag and measure again. What mass changes do you expect? If you have a bag and a balance, do the experiment.

Then, evaluate the protocol you used in point 1.

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