What is the volume of the balloon in the liquid nitrogen?

Question

A balloon filled with helium gas at 20 degrees C occupies 2.91 L at 1.00 atm. The balloon is immersed in liquid nitrogen at -196 degrees C that applies an external pressure of 5.20 atm to it. What is the volume of the balloon in the liquid nitrogen?




(Please Show Work!)

Answer 1

There is a little trick to this question, Alyssia.
We can try to use the ideal gas law to calculate the change in volume.
P1V1 / T1 = P2V2 / T2

Keep in mind that temperature is absolute, not celsius, so T1 is 273.15 + 20 = 293.15 and T2 is 273.15 - 196 = 77.15

Solving for the second volume, we get the equation V2 = V1 * P1/P2 * T2/T1.
Substituting the given values we have V2 = 2.91 * 1/5.2 * 77.15/293.15
Therefor the new volume might be .147 L.

However, the balloon will loose all elasticity before reaching that temperature. Therefor the actual starting and ending pressures of the gas INSIDE the ballon cannot truely have the same ratio as the starting and ending pressures outside the ballon. For a real balloon, the answer is too difficult for me to calculate.

For a theoretical balloon that keeps a constant elasticity, the answer is .147 L.

Answer 2

Is this a trick question?
The rubber of the balloon will rapidly become hard at liquid nitrogen temperatures so the balloon may not shrink in fact it may shatter due to the external pressure.

Answer 3

You can arrange the ideal gas law as PV/T = nR. Since nR is a constant for this situation, you have PV/T = PV/T. Convert the temperatures to Kelvin, and substitute the initial pressure, volume and temperature on the left and the final pressure and temperature on the right and solve for the final volume.

I'll let you solve the question and show your own work, not mine.