Evan's Space

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.

Unsual Behaviour of Less Dense Balloon in Air or Water

Of course if a heavy ball is suspended from the car we all know the ball will move in opposite direction of a decelerating car due to inertia. We are all familiar to this where inertia is applied to a body which is denser than the surrounding medium (air or liquid) which is less dense. That’s why this balloon’s behaviour surprises us!

Let’s assume the water molecules are initially moving at constant speed with the tank before the deceleration.  Due to inertia, when the tank decelerates, the water molecules continue its state of motion forward. Hence the water molecules gush to the right side of the tank, displacing (pushing) the balloon to the left. Hence the balloon moves to the left!

Refer to this Youtube video by Smarter Everyday and you can see the similar experiment of helium balloon (less denser) in the air (denser) of a car.

Marshmallow Hulk in Vacuum Jar

When the pump is switched on and the air in the jar is gradually removed, the pressure in the jar decreases. There will be fewer air molecules per unit volume in the far. Hence rate of collision of the air molecules with one another and with the wall and hulk will be reduced. As pressure P = F/A, the force acting per unit area decreases, the pressure decreases.

In the marshmallow, there are pockets of air at normal atmospheric pressure initially. As the pressure in the jar decreases, the pockets of air in the marshmallow expands due to this pressure difference. Hence the hulk expands and its volume increases.

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

Gas expands and contracts the most

The 3 states of matter –  solid, liquid and gas.

In general, when a body is heated, it expands and volume increases. The mass remains the same. Since density = mass/volume, its density decreases (less dense). For instance, warm air rises as it is less dense. In terms of kinetic theory, the particles will increase in kinetic energy. The average spacing between the particles increases (assuming not in a closed container).

Likewise, when a body is cooled, the opposite occurs. The body contracts and volume decreases. It becomes denser.

Due to the differences in particles arrangement of solid, liquid and gas, each expands by different amount when heated and vice versa. Which expands the most when heated and contracts the most when cooled?

The following demonstration of the ‘Pee Boy’ is a good video to show the concepts.

Explanation:

The tiny hole at the penis is too small for any water to enter on its own. So using thermal transfer in the different states, the following steps are taken:

1. Put the hollow empty boy into the hot water. [air inside the boy expands more than the solid ceramic, hence bubbles are seen coming out of the hole]
2. Put the hollow empty boy now into the cold water. [The air inside contracts and volume decreases. This creates a low pressure and water is then sucked into the boy through the tiny hole]
3. Place the boy on a platform. [The boy is only partially filled with water. The head portion is filled air while the bottom portion is filled with water]
4. Pour hot water over the head. [As the whole boy is heated by the running hot water, the air in the head portion expands much more than the water at the bottom and the solid ceramic of the boy. Hence the air pressure increases and it pushes the water out of the boy]
5. And he pees!!! Quite powerful indeed!

2013 Nov Sci Phy P2 Q9 – Pressure and Moment

Solutions:

(a) (i) A bigger force than F can be obtained due to the level system and hydraulic system.
Level system: Applying principle of moments, the anticlockwise moment by the F is equal to the clockwise moment by the force on piston A (note that the handle is pushing the piston down, but the piston A is pushing on the handle upwards – action = reaction). As the perpendicular distance from F to the pivot is greater than the perpendicular distance of the force by piston to the pivot, the force on the piston A is greater than F at handle.
Hydraulic system: As the pressure transmitted in the liquid is the same, pressure at piston A = pressure at piston B. As P = F/A and area of piston A is smaller than area of piston B, a larger force is obtained in piston B.
Hence these two systems allow the force on piston B to be greater than F at the handle.

(a) (ii) Both liquid and gas molecules are in a continuously random motion. But in liquid, the molecules are closely packed together and able to slide around one another. There is very little empty space between the molecules hence liquid is not compressible. Gas molecules are far apart from one another, hence gas can be easily compressed.

(b) P = F/A = 12000 / 0.060 = 200 000 Pa

(c) Velocity is the vector quantity while speed is a scalar. As the car goes round the bend, the direction of the car changes. Hence velocity is changing even though speed is constant.

Likewise, as the velocity is changing, the car is considered to have an acceleration (not in the sense of increasing speed though).

N2004 P1 Q15 – Pressure P1V1 = P2V2

Water if depth 10 m exerts a pressure equal to atmospheric pressure. An air bubble rises to the surface of a lake which is 20 m deep. When the bubble reaches the surface, its volume is 6 cm3.

What was the volume of the air bubble at the bottom of the lake?

A) 2 cm3          B) 3 cm3         C) 12 cm3      D) 18 cm3

Solutions: Option A
Since the air bubble is enclosed, PV at A is equal to PV at B.

PAVA = PBVB

(Patm + Pwater) VA = (Patm) VB

(10 + 20) VA = 10 x 6

VA = 60 / 30

= 2 cm3

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.