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How does the air temperature affect the size of a balloon?
Answer 1:

The volume that a given parcel of air takes up is dependent on the Ideal Gas Law:


n=number of moles (similar to number of gas atoms)
T = temperature

According to this equation, if Temperature on the right hand side goes up then volume on the left hand side will also have to go up for them to still be equal.

To put this in simpler terms, heat is energy, when you run really fast your body heats up, your body is burning lots of energy and it creates heat. You can think of heat as the byproduct of things moving fast.

If you rub your hands together really quickly, they will be warm too. In a balloon when the temperature increases, the individual gas molecules that you can't see but that make up gases start moving really fast. When the molecules are moving fast under higher temperatures, they run into the sides of the balloon more often making the balloon bigger. Basically a warmer temperature leads to higher energy gas molecules which take up more space since they are moving faster. If you cool down the temperature, the gas molecules slow down and they take up less space.

Hope that helps.

Answer 2:

As temperature increases, so does the volume of the balloon. In fact, they are proportional to each other if pressure does not change. There is a catch though: in order to be able to perform calculations, you need to use what is known as absolute temperature. If you are using degrees Fahrenheit, you need to add 460 to get absolute temperature. These would then be called degrees Rankine (R). For example if the balloon's temperature is 73F, its absolute temperature would be 533R. If you now take the balloon outside where the air temperature is, say, 20F and therefore 480R, and you wait long enough for the air (or helium or any other gas) inside the balloon to reach that same temperature, then the balloon would shrink in the same proportion. Say the balloon's volume was 2 cubic feet when its temperature is 73F. By solving a rule of three (try the math yourself: 533 is to 2 as 480 is to x), you will get a volume of 1.8 cubic feet for the cooled down balloon. Notice that even though this is a rather extreme case where the temperature change is 53F, you only get a 10% reduction in volume. To add insult to injury, since the volume of a balloon is proportional to its diameter raised to the third power, a 10% reduction in volume means an only about 3.5% reduction in diameter which is very hard to notice just by looking at it. In our example, the diameter of the balloon would be about 18 3/4 inches initially and the reduction of diameter would be only about 5/8 inch when it cools down to 20F.

One thing you can do now that winter is approaching is the actual experiment. If you live in California or Florida, maybe you can use your freezer to cool down the balloon instead. Just make sure the balloon is not inflated too tight so the pressure inside the balloon will be about the same before and after. It would be easier to measure the circumference instead of the diameter and, since diameter is proportional to circumference, then the volume of the balloon is proportional to the third power of its circumference. Once you compute the reduction in volume as a percent, simply divide this percent by 3 to get the approximate reduction in diameter also as a percent. If you become proficient in measuring the circumference of a balloon you can even use it instead of a thermometer to measure temperatures inside freezers or in a cold outside winter. Just work your way backward:measure the circumference before and after, then find the percent reduction in circumference, multiply by 3 to get the percent reduction in volume. This will be the same percent reduction in absolute temperature which will allow you to find the absolute temperature and hence the temperature in F. You would still need to know the temperature inside your home of course.

This same procedure will work in the summer where outside temperature will be greater than temperature inside your home (let us hope). Instead of a reduction in temperature, volume, diameter and circumference referred to above, you will have an increase, with the same relationship between percent changes.

Have fun!

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