Volume ratio

[Type text] [Type text] [Type text] _An experiment on the effect of surface area to volume ratio on the rate of osmosis of Solanum tuberosum L._ BACKGROUND A cell needs to perform diffusion in order to survive. Substances, including water, ions, and molecules that are required for cellular activities, can enter and leave cells by a passive process such as diffusion. Diffusion is random movement of molecules in a net direction from a region of higher concentration to a region of lower concentration

Surface Area to Volume Ratio and the Relation to the Rate of Diffusion Aim and Background This is an experiment to examine how the Surface Area / Volume Ratio affects the rate of diffusion and how this relates to the size and shape of living organisms. The surface area to volume ratio in living organisms is very important. Nutrients and oxygen need to diffuse through the cell membrane and into the cells. Most cells are no longer than 1mm in diameter because small cells enable nutrients and oxygen

“paths” from the sides of the body that are capable of releasing this heat particles, and reaching thermal equilibrium faster. This is what happens when a hotter body is subjected to a colder one. Research Question: How does the surface area to volume ratio

small meaning that their surface area is big compared to their volume. These organisms have a large surface area to volume ratio which means that they can obtain different substances by diffusion through their relatively large plasma membrane. The substances have to diffuse only short distances so they can diffuse at a faster rate and meet the organism’s needs. Multi-cellular organisms have a much smaller surface area to volume ratio. Many of their cells are not in direct contact with their surroundings

Surface area / Volume ratio Experiment Introduction: The surface area to volume ratio in living organisms is very important. Nutrients and oxygen need to diffuse through the cell membrane and into the cells. Most cells are no longer than 1mm in diameter because small cells enable nutrients and oxygen to diffuse into the cell quickly and allow waste to diffuse out of the cell quickly. If the cells were any bigger than this then it would take too long for the nutrients and oxygen to diffuse into

area to volume ratio. Many of their cells are not in contact with their surroundings so they con not only rely on diffusion to supply all their organs with oxygen and nutrients, as the distance from their surface to all cells is too fare. We are multi cellular and have special surface for gaseous exchange and for obtaining nutrients. However Single celled organisms are small, which means that their surface area is large compared with their volume; they have a large surface area to volume ratio. Therefore

Formula Volume Formula (cube 2 S.A. 6  s 3 V s  sphere 2 S.A. 4  r 4 3 3 V r   cylinder 2 S.A. 2 2     r Rh 2 V r h. Cube: s/6 Sphere: 0.333333r Cylinder: 6.283185h+6.283185r Volume to surface ratio: = s^3 / 6s^2 = s/6. Cube shaped sphere shaped cylinder shaped Volume 16 16 16 Surface area 38.097625 30.706532 35.207969 Volume/S.A Ratio .419973 .521061 .454442 S.A/Volume Ratio

change in acceleration) and the volume (how much space it takes up). To calculate density the formula used is, F = m/v. Mass is measured in g, volume is measured in mL, and density is measured in cm³. The hypothesis, “If the density of a small amount of a substance is compared to the density of a large amount of the same substance then they densities will be the same because the ratio between mass and volume remains the same,” was proven true as when mass is increased, volume increased as well at the same

size and shape on the ratio of surface area to volume. Complete the following exercises. Part 1 Let’s investigate how size influences surface are to volume ratios. Let’s assume we have a cube with a volume of 1cm3. Each side of the block equals 1cm. Picture a die for each cube. 1. How many sides does a cube have? 2. Let’s use 1 block to represent a small cubed organism and 8 blocks to construct a large cubed organism. Complete the following table. Total Volume Total surface area

area to volume ratio on the rate of diffusion is investigated. Four agars containing phenolphthalein, an indicator that turns colorless in acid, each ranging in size were used. In this experiment, the agar cubes represent cells and the time it took for cubes to turn colorless shows the rate of diffusion. Next, the cubes were placed in a beaker and just enough hydrochloric acid was added to cover them. Then, the time it took them to turn colorless was measured. It was observed that when volume increases