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).
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
T = Tstp = 273.15 K
P = Pstp = 1 atm and
n = 1 mol gas
R = 0.08206 L-atm/mol-K
V = Vstp = (n × R × Tstp)/Pstp
= ((1 mol) × (0.08206 L-atm/mol-K))/(1 atm)
= 22.414 L / mol
Question 1: What is the volume of oxygen that, as 2.50 mol, exerts a pressure of 1500 mm mercury at 20 degrees C?
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?
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?
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?
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?
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 Vm.
= (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?
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
Question 7: Where Argon gas occupies 50 L determines the number of moles and mass of the argon sample?
One mole of gas = 22.4L.
= 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?
One mole of gas = 22.4 L.
= 1.25 mol × 22.4 L/mol
= 28.4 L.