# 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 × TWhere as, P = Pressure

V = Volume

n = no. of moles

R = molar gas constant

T = Temperature

At STPT = T

_{stp}= 273.15 KP = P

_{stp}= 1 atm andn = 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 N

_{2}O = ((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 N

_{2}O 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 co

_{2}behaves ideally 5 moles,= 22.4 L × 5 = 112 L.

**Question 4: The gram equivalent volume of O _{2} 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 O

_{3}using the formulan = m/M, where,n = mole =?

m = mass = 55 g O

_{3}M = molar mass = 3×15.999 g/mol O

_{3}= 47.997 g/moln = (55 g O

_{3}/47.997 g/mol) = 1.145 mol O_{3}Calculate volume of O

_{3}by multiplying mol O_{3}by the V_{m}.= (1.145 mol O

_{3}×22.414 L/mol) = 25 LThe volume of 55 g of O

_{3}at STP is25 L.

**Question 6: What is the volume of gas in STP in cm ^{3}?**

**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.