The molar volume of a gas is
equal to the volume occupied by the 1 mole of a gas at given conditions
of temperature and pressure. The molar volume can be obtained by
dividing the molar mass by the density of the gas at given temperature.
Molar volume = molar mass / density
The unit of molar volume is M3 / mole or dm3/mole
PV = nRT
V/n = RT/P plug in the values of R = 8.31 joule / mole / Kelvin, let the temperature = 273 Kelvin and the pressure as 1 Pascal = 1.01*105 N/m2. These are the STP conditions.
VMOLAR = (8.31*273/1.01* 10 5)=22.4 liters approximately.
So the molar volume of an ideal gas at ST P is equal to 22.4 liters
The molar volume of a real gas can be obtained by using the vanderwaal’s equation:
(P+a/V2) (V-b) = nRT
The a and b are the constant and different for a gas specific.
By approximation and Avogadro’s’ law the volume of 1 mole of a gas at STP is 22.4 liter invariably with the nature and type of a gas. Then applying the gas equation we can find the volume of the gas at any temperature.
Molar volume = molar mass / density
The unit of molar volume is M3 / mole or dm3/mole
Introduction to molar volume of a gas :
The volume of an ideal gas can be obtained by the ideal gas equationPV = nRT
V/n = RT/P plug in the values of R = 8.31 joule / mole / Kelvin, let the temperature = 273 Kelvin and the pressure as 1 Pascal = 1.01*105 N/m2. These are the STP conditions.
VMOLAR = (8.31*273/1.01* 10 5)=22.4 liters approximately.
So the molar volume of an ideal gas at ST P is equal to 22.4 liters
The molar volume of a real gas can be obtained by using the vanderwaal’s equation:
(P+a/V2) (V-b) = nRT
The a and b are the constant and different for a gas specific.
By approximation and Avogadro’s’ law the volume of 1 mole of a gas at STP is 22.4 liter invariably with the nature and type of a gas. Then applying the gas equation we can find the volume of the gas at any temperature.
Molar Volume of a Gas Ilustration
Example: find the molar volume of CO2 at 2 atm pressure and 27 degree C.SOLUTION:
The molar volume of CO2 at STP is 22.4 liters therefore applying the followings
P1=1 atm P2 =2 atm
V1= 22.4L V2=??
T1= 273 K T2= (27+273) = 300 K
APPLYING THE EQAUITON
P1V1/T1 =P2V2/T2
V2 = P1T2V1/P2T1
= 1*300*22.4 /(2*273) = 12.30 liters
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