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Chemistry

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Suppose it is a hot summer day (40.0 degrees C = 104.0 degrees F), and you have left a can of hairspray outside so the gas within it is now at the same temperature of its surroundings. As part of your chemistry lab class, you have to build a safe and effective potato gun to share with the local elementary school, and you must be sure that the combustion chamber is well insulated and will not harm (burn) the shooter. You spray the warm hairspray into the 3.50 L chamber on your gun. You seal the cap, then ignite it, and the resulting gas expands to 13.0 L as the potato is shot out. What is the final temperature, in Celsius, inside the combustion chamber? What is this temperature in Fahrenheit? [SHOW ALL WORK TO RECEIVE CREDIT]

  • Chemistry - ,

    This is ridiculous. Is your teacher a chem teacher? This is not a gas expansion problem, it is a chemicalreaction that makes more gas.


    XXXX ignites (meaning oxygen was there), and more hot gas (CO2, steam) is created. You might have had a few moles of hairspray solvent, but in the chem reaction, many more moles of product is created, heated, and it expands to the final volume.

    Point: the initial volume of hairspray undergoes a chem reaction, making more gas, hot, which expands.

    I am wondering about the person who give this gas problem: it is not a thermal expansion problem, as much as a chem reaction problem. The expansion to 13L is not the same hairspray "gas" as went into the chamber.

    Hairspray once had butane as the gas in the can, but now it is mainly ether and ethylalcohol which are flammable. These flammable compounds, which are 40 percent of the mass of hairspray, ignite, and chemically react to make more moles of gas at high temperature, which then increases the pressure and sends a spud flying.

  • Chemistry - ,

    Simple problem and I think this guy is over thinking it. It's an intro to Chem class. Anyway you have initial temp/initial volume. Unknown secondary temp/known secondary volume.

    Charle's law is V1/T1 = V2/T2. I'm sure you can figure it out from here...

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