PROBLEM 127P

In a common classroom demonstration, a balloon is filled with air and submerged into liquid nitrogen. The balloon contracts as the gases within the balloon cool. Suppose the balloon initially contains 2.95 L of air at 25.0 °C and a pressure of 0.998 atm. Calculate the expected volume of the balloon upon cooling to —196 °C (the boiling point of liquid nitrogen). When the demonstration is carried out, the actual volume of the balloon decreases to 0.61 L. How well does the observed volume of the balloon compare to your calculated value? Can you explain the difference?

Solution 127P

This has to do with Charles Law that states volume is proportional to temperature (if one goes up the other one does and vice versa).

The formula you must use is

v1/t1 = v2/t2

* v = volume

* t = temperature

Now substitute in the values... so...

2.94 L / 296 K = v2 / 77 K

v2 = (2.94 L / 296 K )(77 K)

v2 = 0.76 L

Observed volume of the balloon is the actual volume that occurred. Calculated volume of the balloon is what was expected based on math. Often these are not the same because little errors in experimental design.