Balloon at reduced pressure

A partially inflated balloon is placed inside a sealed container as air is pumped out of the by syringe. As the pressure of the container is reduced, the volume of the balloon increases. The pressure inside the balloon decreases.

Note that the pressure inside the balloon is not equal to the pressure of the container. The number of air molecules in the balloon is fixed. As the pressure of the container decreases, this creates a pressure difference inside and outside the balloon. This causes the balloon to expand and its volume to increase. The number of air molecules per unit volume inside the balloon decreases, hence pressure inside the balloon decreases.

But as the balloon is elastic, the wall of the balloon is stretched. Hence the pressure of the balloon will be greater than the pressure of the container. Refer to the video below.

 

Another P1V1 = P2V2 question

Solution: Option C

When to use the concept PV = constant and P1V1 = P2V2 to solve?

When temperature is constant (for o-level), when a fixed mass of gas (fixed number of air molecules) is compressed in a closed system (e.g. piston), the volume V decreases and pressure P increases, and vice versa.

But when you multiply pressure and volume, PV, it is always a constant.

PV = constant

Hence we can always equate the PV of the first scenario = to the PV of the second scenario, provided there is no addition or removal of air molecules from the system.

Hence, you have P1V1 = P2V2

The followings are 4 different questions which require this concept to solve. Do revise them.

Solutions: Option D (refer to the worked solutions below)

Solutions: Option D

Solutions: A

4)

Solutions:

 

5) 

Solution: Option C

Sketching P-V, P-1/V and PV-P graph

Pressure of a fixed mass of gas, P, is inversely proportional to the volume of the gas, V, when the temperature is held constant.

P inversely proportional to V

P = k/V , where k is a constant.

PV = k

With these concepts, you should be able to sketch the various graph. Do not memorise them. Use your concepts from Maths to help you to recall instead.

N2008P1Q15 – Kinetic Model of Matter

Gas inside a cylinder is heated slowly to a higher temperature. The pressure inside the cylinder remains constant as the piston moves outwards.

How do the speed of the gas molecules and their rate of collision with the piston compare with their initial values at the lower temperature?

Solutions:

In short:  Temperature increases, Kinetic Energy increases, Rate of Collision decreases, Average Force on wall increases, Pressure constant.

As temperature increases, the speed of molecules increases, the kinetic energy of the air molecules increases.

As piston is free to move, it will move to the right such that the pressure remains constant (equal to atmospheric pressure outside). As the piston moves to the right, the volume inside the piston increases. 

Surface area in which the air molecules collide increases.

The rate of collision decreases as the number of molecules remains constant. With higher KE of molecules, the molecules will collide the wall with greater force. Though rate of collision decreases, with each collision having greater impact force, the average force acting on the wall of piston increases.

Since P = F / A, with greater force F, over a bigger area A, the pressure P remains constant. (Compared with previously,  smaller F over smaller A, but P constant)  

Misconception: Many think that the rate of collision remains the same, which is wrong. Apparently the effect of volume increases is more significant, hence rate of collision decreases, even though they collide with greater impact force. Hence overall force on wall still increases. If the speed of the molecules increases but the pressure remains constant, then the molecules must collide less frequently. If the rate of collision stayed the same, the pressure would increase.  If the rate of collision increased, the pressure would increase even more.