Question

A particular balloon can be stretched to a maximum surface area of 1257 cm2. The balloon is filled with 3.1 L of helium gas at a pressure of 754 torr and a temperature of 294 K . The balloon is then allowed to rise in the atmosphere. Assume an atmospheric temperature of 273 K and determine at what pressure the balloon will burst. (Assume the balloon to be in the shape of a sphere)

Answer 1

Answer:

The ballon will brust at

Pmax = 518 Torr ≈ 0.687 Atm

Explanation:

Hello!

To solve this problem we are going to use the ideal gass law

PV = nRT

Where n (number of moles) and R are constants (in the present case)

Therefore, we can relate to thermodynamic states with their respective pressure, volume and temperature.

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$ --- (*)

Our initial state is:

P1 = 754 torr

V1 = 3.1 L

T1 = 294 K

If we consider the final state at which the ballon will explode, then:

P2 = Pmax

V2 = Vmax

T2 = 273 K

We also know that the maximum surface area is: 1257 cm^2

If we consider a spherical ballon, we can obtain the maximum radius:

$R_{max} = \sqrt{\frac{A_{max}}{4 \pi}}$

Rmax = 10.001 cm

Therefore, the max volume will be:

$V_{max} = \frac{4}{3} \pi R_{max}^3$

Vmax = 4 190.05 cm^3 = 4.19 L

Now, from (*)

$P_{max} = P_1 \frac{V_1T_2}{V_2T_1}$

Therefore:

Pmax= P1 * (0.687)

That is:

Pmax = 518 Torr