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# Specific Gravity Formula

• Last Updated : 01 Feb, 2022

Relative gravity or Specific gravity can be defined as the density ratio of a substance to that of water at a specific temperature. Therefore, Specific gravity tells us if an object will sink or float.

It’s important to remember that temperature is a big factor in determining a substance’s specific gravity. Specific gravity is thus a property of a substance at a specific temperature. Furthermore, pressure has an effect on a substance’s specific gravity.

### What is Specific Gravity?

The specific gravity of an object is the ratio of its density to that of a reference substance.

The specific gravity can indicate whether an object will sink or float in our reference substance dependent on its value. Water is commonly used as a reference substance since it has a density of 1 gram per millilitre or 1 gram per cubic centimetre. Moreover, there are many factors that determine whether an object will float or sink.

### Formula for Specific Gravity

The specific gravity formula is defined as the ratio of the density of an object to the density of water, with water as the reference substance.

Mathematically:

Specific Gravity = Density of Water at 4°C / Density of the given substance

​= ​ρsubstance​​​ / ρH2​O

where

ρsubstance ​is the Density of the given substance and

ρH2​O​ is the Density of Water at 4°C​

Unit of Specific Gravity: Since it is a ratio of two densities, therefore it is Unitless.

Dimension of Specific Gravity: As it is unitless, specific gravity is Dimensionless also.

### Factors affecting Specific Gravity

• The temperature and pressure of the environment or surroundings have an impact on specific gravity levels.
• Specific gravity is usually measured at 40 degrees Celsius. At 40 degrees Celsius, the specific gravity of water is one. It is critical to remember that as the temperature changes, this value will vary as well.
• A substance can melt (turn from solid to liquid), evaporate (turn from liquid to gas), or sublimate (turn from solid to gas) when the temperature rises (solid to gas directly). This will modify the density of the substance because solids are the densest of the three and gases have the least density, while the density of the liquid is in the middle. As a result, specific gravity changes as the temperature fluctuates.

### Uses of Specific Gravity

1. The value of gems can be compared to the ideal settings to determine their purity. Hence, it is used by gemologists to discriminate between similar gems.
2. Specific gravity values are also used by geologists and mineralogists to determine the mineral content of any rock.
3. Urinalysis professionals evaluate the contents of urine by measuring its specific gravity.
4. Monitoring the progress of reactions as well as the concentration of solutions.
5. Battery fluid and antifreeze are being tested using specific gravity.

### Sample Problems

Problem 1: What is the specific gravity of a substance with a mass of 10 kg and volume 2 m3, with respect to the reference density of 50 kg/m3?

Solution:

Given that –

Mass of the substance, m = 10 kg

Volume, V = 2 m3

This implies, the density of the substance, ρ = m/V = 10 kg/2 m3 = 5 kg/m3

And the reference density, ρ’ = 50 kg/m3

The formula to calculate the specific gravity of a substance is:

Specific Gravity = Density of the given substance / Reference Density = ρ/ρ’

= 5 kg/m3 / 50 kg/m3

= 0.1

Problem 2: A liquid has a mass of 20 gm and the volume of the water (reference material) is 2 mL. Find the specific gravity of the substance?

Solution:

Given that,

m = 20 g

V = 2 mL

ρ = 1 g/mL

Then the density of the substance is,

ρ’ = 20 g/2 mL = 10 g/mL

Therefore,

Specific Gravity = ρsubstance/ρH2O = 10g/mL / 1g/mL = 10

Hence, the density of the substance is 10 g/mL and the specific gravity is 10.

Question 3: Determine the specific gravity of gas with mass 5 kg and volume 1 m3, with respect to the reference density 10 kg/m3?

Given that –

Mass of the substance, m = 5 kg

Volume, V = 1 m3

This implies, the density of the substance, ρ = m/V = 5 kg/1 m3 = 5 kg/m3

And the reference density, ρ’ = 10 kg/m3

The formula to calculate the specific gravity of a substance is:

Specific Gravity = Density of the given substance / Reference Density = ρ/ρ’

= 5 kg/m3 / 10 kg/m3

= 0.5

Question 4: A gas has a mass of 10 gm and the volume of the water (reference material) is 10 mL. Find the specific gravity of the substance?

Solution:

Given that,

m = 10 g

V = 10 mL

ρ = 1 g/mL

Then the density of the substance is,

ρ’ = 10 g/10 mL = 1 g/mL

Therefore,

Specific Gravity = ρsubstance/ρH2O = 1 g/mL / 1g/mL = 1

Hence, the density of the substance is 1 g/mL and the specific gravity is 1.

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