A balloon has a pinhole 10 micrometer diameter. It is filled with O2 at pressure

Question

A balloon has a pinhole 10 micrometer diameter. It is filled with O2 at pressure 2 * 10^4 Pa, and T=298 K. a) How many molecules per second escape? b) The ballon is a sphere 1m in diameter. How long would it take for 1% of the gas to escape? c) Consider the same balloon, with N2 added at the same temperature, P= 8 * 10^4 Pa. How would this change the number of O2 molecules escaping(the partial pressure of O2, remains 2 * 10^4 Pa)? How would this change the time to the escape of 1% of the total gas?

Explanation / Answer

Ans : (a) Suppose we have a container that contains gas at initial pressure P and temperature T. A small hole of area A is made in the container. Suppose that the average x component of the velocity of those molecules that are moving towards the hole is vx. Assuming elastic collisions, each molecule has its x momentum of mvx reversed, so that the momentum change per sec as a result of collisions with an area A is the force, so

F= N(2mvx)/ t = P*A or N= (PA t)/ 2mvx

Using Kinetic theory of gases we know that vx (rms) = RT/M

so we can use the rms of vx as an estimate for vx

If the area A is now made into the hole, then assuming that there is a vacuum on the other side of the hole, all the molecules that would have collided with the hole will now escape from the container, and no molecules will pass into the container.

N= (PA t)/ 2mvx

for t =1 sec, N= (PA)/ 2mvx

vx = RT/M = (8.314 x 298)/32 x 10-3 = 278.2 m/s

Thus N= (PA)/ 2mvx = [2 x 10-4 x 3.14 x (10-5/2)2]

                                    2 x 278.2 x (32 x 10-3/6.023 x 1023)

Upon solving, we get N = 5.11 x 108 molecules

(b) Consider N= (PA t)/ 2mvx

N/ t = - (PA)/ 2mvx (the minus sign just indicates that N decreases)

            = -(PA)/(2mkT/m) = -(NkTA)/(2mVkT/m)= - (NAkT/m)/2V

N/ t =- (NTkT/m)/2V

Treating this as a differential equation, we can solve it to get

If 1% gas escapes i.e N(t)/ N(0) = .99, then

0.99= exp (-AtkT/2V m)

Solving we get t = 3.85 x 106 s

(c) Upon addition of N2, the partial pressure of oxygen doesn’t change and thus the number of molecules of oxygen striking the pinhole per sec remains unchanged. Hence the number of molecules of oxygen escaping the balloon will not change.

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