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.

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.

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. Question 01 Solutions: Option D (refer to the worked solutions below) Question 02 Solutions: Option D Question 03 Solutions: A Question 04 Question 05 Solution: Option C

The video below demonstrates that water will find its own level. An question requires this concept: An apartment block receives water from a nearby reservoir. A pump is necessary to lift the water into a storage tank at the top of the building. The gravitational field strength is 10 N / kg. How much energy does the pump supply to lift each kilogram of water into the tank? Solutions: 100J Water will always find its own level. Hence with or without the pump, the water level in the pipe in the building will be of the same level as the reservoir. Hence, the pump is only needed to pump water up a height of 10m only (instead of the 50m). Energy, E = mgh = 1 kg x 10 N/kg x 10m =  100 J

Answer: Option B Since water will naturally reach 40m high from the bottom of the apartment, you only need to pump the water up by a further 10 m. Energy needed = GPE = mgh = 1 x 10 x 10 = 100 J View this video to understand that when both ends of the tube are exposed to atmospheric pressure, the water levels on each end will be level.

At sea level (where most of us are), the standard atmospheric pressure is about 101325 Pa. The boiling point of water is at  100 oC which we are familiar with. But as you climbed up e.g. Mount Everest at 8,848 m, the pressure is low and the boiling point of the water is about 71oC. So that’s the hottest cofe you can have on top of the cold mountain! Hence as the pressure decreases, the boiling point of the water decreases. As with lower pressure, the water molecules requires lesser energy to break the intermolecular forces to escape into the atmosphere, hence boiling point is lower. This video shows the same effect. Using the syringe, the air is pumped out of the container to reduce the pressure. The water at 75  oC , (below the usual boiling point of 100 oC) will start to boil and you can observe the bubbles forming!