STP Formula

Standard temperature is defined as the freezing point of pure water at sea level, 0˚C (zero degrees Celsius), or 32˚F (32 degrees Fahrenheit) or 273˚K (273.15 degrees kelvin). Standard pressure is defined as a unit of pressure that will support a column of mercury 760 mm high at sea level and 0 degrees centigrade. The STP is defined as 0˚C and 1 atm. The volume of a gas depends on several factors. The gas’s temperature, the gas’s pressure, and the amount of gas (number of moles).

STP Formula

The ideal gas law states that the volume occupied by a gas depends on the amount of substance (gas), temperature, and pressure. Standard temperature and pressure are 0 degrees Celsius and 1 atmosphere of pressure. Parameters of gases are significant for calculations in chemistry and physics and are usually calculated at STP.

By using the law of Ideal gas, 

P × V = n × R × T

Where as, P = Pressure

V = Volume

n = no. of moles

R = molar gas constant

T = Temperature

At STP 

T = T_stp = 273.15 K

P = P_stp = 1 atm and

n = 1 mol gas

R = 0.08206 L-atm/mol-K

V = V_stp = (n × R × T_stp)/P_stp

= ((1 mol) × (0.08206 L-atm/mol-K))/(1 atm)

= 22.414 L / mol

Sample Questions

Question 1:  What is the volume of oxygen that, as 2.50 mol, exerts a pressure of 1500 mm mercury at 20 degrees C?

Solution: 

We know that from the definition, 1∙atm will support a column of mercury that is 760∙mm high. We use Kosher units,

i.e., P = (1500 mm. Hg)/(760 mm. atm^-1)

= 1.97 atm.

Then, we solve for volume in the Ideal Gas Equation,

V = (n × R × T)/P

= ((1.50∙mol) × (0.0821∙L∙atm/K∙mol) × (293∙K))/(1.97∙atm)

= 30.5 L.

Question 2: Is the volume of 25 g dinitrogen monoxide gas at STP?

Solution:

Nitrogen has a molar mass of 14.0067 g/mol, and oxygen has a molar mass of 15.999 g/mol. 

 Molar mass of N₂O = ((2 × 14.0067) + 15.999) g/mol = 44.0124 g/mol. 

The molar volume of gas at STP is 22.4 L/mol. 

Thus, the volume of 20.5 g of N₂O is,

= ((22.4 L/mol) × (25 g)) /(44.0124 g/mol)

= 12.7 L

Question 3: What is the volume of 5 moles of Carbon dioxide gas at STP?

Solution: 

An ideal gas at STP occupies 22.4 L.

Assuming co₂ behaves ideally 5 moles, 

= 22.4 L × 5 = 112 L.

Question 4: The gram equivalent volume of O₂ at STP?

Solution:

The gram equivalent mass of oxygen is 16.

32 g of Oxygen occupies 22.4 L at STP.

So, applying simple unitary method, the gram equivalent volume of oxygen will be 11.2 L.

Question 5: Is the volume of 55g of Ozone or trioxide at STP?

Solution:  

The volume of one mole of a gas is 22.414 L/mol.

You need to calculate the number of moles in 55 g O₃ using the formula n = m/M, where,

n = mole =?

m = mass = 55 g O₃

M = molar mass = 3×15.999 g/mol O₃ = 47.997 g/mol

n = (55 g O₃/47.997 g/mol) = 1.145 mol O₃

Calculate volume of O₃ by multiplying mol O₃ by the V_m.

= (1.145 mol O₃×22.414 L/mol) = 25 L

The volume of 55 g of O₃ at STP is 25 L.

Question 6: What is the volume of gas in STP in cm³?

Solution:

1 mol of gas has a volume = 22.4 L.

 We know that there are 1000cm³ in 1L.

The volume of 1mol gas at STP 

= 22.4L × 1000cm³/L 

= 22,400cm³

Question 7: Where Argon gas occupies 50 L determines the number of moles and mass of the argon sample?

Solution:

One mole of gas = 22.4L.

Therefore,

= 50 L.Ar/22.4L

= 2.23 mol.Ar

We know that mass of argon = 39.95

= 2.23 mol × (39.95/ 1 mol.Ar)

= 89 g.Ar.

Question 8: What volume will 1.25 moles of Helium gas occupy at STP?

Solution:

One mole of gas = 22.4 L.

Therefore, 

= 1.25 mol × 22.4 L/mol

= 28.4 L.