Conservation of mass
Definition
Conservation of mass
The principle that the total mass of a closed system stays the same when its material changes shape, state or volume.
Definition
Closed system
A system that no matter can enter or leave.
- Particles cannot be created or destroyed when a material is squashed, stretched, heated, cooled, melted or frozen.
- If no material enters or leaves, the number of particles stays the same, so the mass stays the same.
- This idea is called conservation of mass, and it applies to any closed system.
- A sealed syringe of air, a block of ice melting in a sealed bag and a lump of clay being reshaped are all examples of closed systems.
- Many changes still alter the volume of a material, because the particles move closer together or further apart.
- A change in volume with no change in mass must therefore produce a change in density.
How volume affects density
Definition
Inverse proportion
A relationship between two quantities in which one halves when the other doubles, because their product stays constant.
- For a fixed mass, the density equation ρ=mV\rho = \dfrac{m}{V}ρ=Vm shows that density and volume are inversely proportional.
- If the same mass takes up a larger volume, the density decreases, because the mass is divided by a bigger number.
- If the same mass takes up a smaller volume, the density increases, because the mass is divided by a smaller number.
- Doubling the volume of a fixed mass halves its density, and halving the volume doubles its density.
- Since the mass is the same before and after the change, ρ1V1=ρ2V2\rho_1 V_1 = \rho_2 V_2ρ1V1=ρ2V2, where the subscripts 111 and 222 label the values before and after.
- Changing only the shape of a solid, such as rolling clay into a snake, keeps both its mass and its volume the same, so its density does not change.
- Cutting a block in half halves both its mass and its volume, so each piece still has the same density as the whole block.
Examples of conserved mass
- When a gas is compressed in a sealed syringe, its particles are pushed closer together, so its volume falls and its density rises.
- When air in a hot air balloon is heated, it expands, so the same mass fills a larger volume and its density falls.
- When water freezes, its volume increases by about 9%9\%9%, so the ice formed is less dense than the water it came from.
- When most other liquids freeze, their volume decreases slightly, so the solid formed is slightly denser than the liquid.
- When water boils, the same mass of water becomes steam with a volume about 170017001700 times larger, so its density falls by the same factor.
- In every case, the mass of the material before the change is used as the mass after the change.
Example
Water freezing in a bottle
- A bottle holds 900 cm3900\ \text{cm}^3900 cm3 of water of density 1.00 g/cm31.00\ \text{g/cm}^31.00 g/cm3, and the water freezes into ice of density 0.92 g/cm30.92\ \text{g/cm}^30.92 g/cm3.
- Find the mass of the water: m=ρV=1.00×900=900 gm = \rho V = 1.00 \times 900 = 900\ \text{g}m=ρV=1.00×900=900 g.
- The bottle is sealed, so the mass of the ice is also 900 g900\ \text{g}900 g.
- Find the volume of the ice: V=mρ=9000.92=978 cm3V = \dfrac{m}{\rho} = \dfrac{900}{0.92} = 978\ \text{cm}^3V=ρm=0.92900=978 cm3.
- The volume increases by about 80 cm380\ \text{cm}^380 cm3, which is why a full glass bottle of water can crack in a freezer.
Example
Compressing air in a syringe
- A sealed syringe contains 0.060 g0.060\ \text{g}0.060 g of air in a volume of 50 cm350\ \text{cm}^350 cm3.
- The density before compression is ρ=0.06050=0.0012 g/cm3\rho = \dfrac{0.060}{50} = 0.0012\ \text{g/cm}^3ρ=500.060=0.0012 g/cm3.
- The plunger is pushed in until the volume is 20 cm320\ \text{cm}^320 cm3, and no air escapes, so the mass is still 0.060 g0.060\ \text{g}0.060 g.
- The density after compression is ρ=0.06020=0.0030 g/cm3\rho = \dfrac{0.060}{20} = 0.0030\ \text{g/cm}^3ρ=200.060=0.0030 g/cm3.
- The volume has been divided by 2.52.52.5, so the density has been multiplied by 2.52.52.5.
Example
Heating air in a balloon
- Air in a balloon has a density of 1.20 kg/m31.20\ \text{kg/m}^31.20 kg/m3, and heating it increases its volume by 25%25\%25% with no air escaping.
- The new volume is 1.251.251.25 times the original volume, so ρ2=ρ1V1V2=1.201.25\rho_2 = \dfrac{\rho_1 V_1}{V_2} = \dfrac{1.20}{1.25}ρ2=V2ρ1V1=1.251.20.
- The new density is ρ2=0.96 kg/m3\rho_2 = 0.96\ \text{kg/m}^3ρ2=0.96 kg/m3.
- The heated air is less dense than the cooler air outside, which is what allows a hot air balloon to rise.
Exam technique
Changes where mass is conserved
- Look for words such as sealed, no material is lost or the same object, which tell you to use the same mass before and after the change.
- Work out the mass first from the starting density and volume, then use it with the new volume or new density.
- State the direction of the change in words as well as numbers, for example the volume increases so the density decreases.
- Keep the units consistent throughout, using either g\text{g}g with cm3\text{cm}^3cm3 or kg\text{kg}kg with m3\text{m}^3m3.
Common Mistake
- Do not assume that a change in volume means a change in mass, because the number of particles stays the same in a closed system.
- Do not say that the density of a piece changes when a block is cut, because the mass and volume fall by the same factor.
- Do not say that particles expand when a material is heated, because it is the spaces between the particles that increase.
- Do not use a new mass after a change of state unless the question says that material has been added or lost.
Self review
- State what is meant by conservation of mass.
- Describe what happens to the density of a fixed mass of gas when its volume is halved.
- Explain why cutting a metal block in half does not change its density.
- Explain why a sealed bottle full of water can crack when the water freezes.
- Calculate the new density of 0.080 g0.080\ \text{g}0.080 g of air compressed from 40 cm340\ \text{cm}^340 cm3 to 10 cm310\ \text{cm}^310 cm3.
