A kilogram is a measure of the mass of a substance used in the international system of units of SI, and a liter is a measure of volume that is not included in this system. The characteristics of physical bodies measured in these units are interconnected by a ratio in which one more parameter is involved - the density of matter. Knowing two of the three parameters - for example, mass and density - calculating the third - volume - will not be difficult.

## Instructions

### Step 1

Start from the general formula connecting mass (m), density (p) and volume (V): V = m / p. Suppose, in the initial conditions, the mass of liquid helium is given, equal to 100 kg, and it is proposed to calculate its volume at standard atmospheric pressure. The density of this substance is 130kg / m³, so 100kg will correspond to a volume approximately equal to 100 / 130≈0, 7692307692307692m³.

### Step 2

Convert the units of the metric system in which the result of the calculation was obtained to liters. In SI, cubic meters are used to measure volume, and one liter holds one cubic decimeter, so increase the value obtained a thousand times - so many cubic decimeters make up each cubic meter. In the example used, the answer should be a value equal to 769, 2307692307692l.

### Step 3

When solving practical problems, take into account the change in the density of the substance during heating. In various reference tables, the density of liquids is listed together with an indication of the measurement conditions, including temperature. And in various regulatory documents, correction factors are indicated separately for the summer and winter periods. For example, for diesel fuel, the summer correction factor is 1.03, and the winter one is 1.045.

### Step 4

If you determine by kilograms a liter volume of bulk solids is required, take into account also the heterogeneity of the material. For example, a one-liter barrel of sand contains not only this substance, but also a certain amount of air between individual grains of sand. This amount depends on the size of the fraction (particle size) constituting the bulk material. In addition, easily deformable substances can be compacted, thereby increasing the average density. Therefore, for example, the weight of cement grade M500 in a column-liter barrel may not correspond to the calculations made on the basis of the tabular density of this substance.