There is no such thing as a truly ideal gas; ideal “perfect’ conditions do not exist. However through numerous assumptions and generalizations, scientists derived a hypothetical equation of how gasses would behave without the confusion of inter-molecular forces, by constraining their definition to the Kinetic-Molecular Theory of Gases. This approximation allows us to predict within a certain degree of accuracy, the way a gas will behave in our design.
Ideal Gas Law using the Universal Gas Constant
Ideal Gas Law [using the Universal Gas Constant] shows the relationship of Pressure, Volume, and Temperature, within all Ideal Gases:
$latex \boxed{PV = n\bar{R}T}&s=3$
Properties (ISA standard conditions at Sea Level):
- p = Absolute pressure (101325)[Pa]
- V = Volume (1)[m^3]
- n = number of moles (42.2925)[mol]
- $latex \bar{R}$ = Universal Gas Constant (8.314462)[J/mol K]
- T = Absolute temperature (288.15)[K]
Example: reorder the equation to solve for moles.
$latex \Huge pV = n\bar{R}T$ becomes $latex \frac{pV}{\bar{R}T} = n &s=2$
$latex \boxed{ \frac{ 101325 _{[Pa]} \times 1 _{[m^3]} }{8.314462 _{[J/mol K]} \times 288.15 _{[K]} } = 42.2925 _{[mol]} }$
(101325[Pa] x 1 [m^3]) / (8.314462 [J/mol K] x 288.15 [K]) = 42.2925 [mol]
So a cubic meter of a gas at Sea Level will always contain 42.2925 moles, regardless of what gas fills that space. Notice that Universal Gas Constant version of the Ideal Gas Law is dependent on molarity (mol/m^3), as we do not know what gas is filling that space. From the resulting moles we can now calculate the mass of a particular substance within that volume and change the equation to one of molality (mass dependent).
Determining the Specific Gas Properties
In order to use the Ideal Gas Law within the context of a known gas, we need to calculate the two dependent properties:
- Molar Mass of the gas and the total mass that is contained in the specified volume (m)
- Specific Gas Constant (R)
We can then calculate the density($latex \rho$) of the gas which will simplify the equation further:
Determine the Molar Mass and Total Mass of the Gas
Find a trusted source for referencing Molar mass. Engineering Toolbox did a beautiful job detailing the calculation of the molar mass of dry air, which is one of the gasses I am studying in the Damn Turbofan project.
Molar mass of dry air = 28.97 (or 28.97 g/mol)
Multiply the quantity of moles we initially calculated, and multiply it by the mass/mol of the known gas.
moles x molar mass = mass of specific gas
$latex \boxed{ 42.29119 _{[mol]} \times 28.97 _{[g/mol]} = 1225.21 g \quad or \quad 1.22521 kg} $
42.29119 [mol] x 28.97 [g/mol] = 1225.21 g or 1.22521 kg
Since we knew the volume of gas previously, we can also calculate the density:
$latex \rho$ = m / V
$latex \boxed{ \frac{ 1.22521 _{[kg]} }{ 1 _{[m^3]} } = 1.22521 kg/m^3 }$
(1.22521 [kg]) / (1 [m^3]) = 1.22521 kg/m^3
Calculate the Specific Gas Constant
Change the Universal Gas Constant from a context of molarity to that of mass. We can do this by dividing the Universal Gas Constant by the molar mass of the known gas.
$latex \boxed{ \frac{ 8.314462 _{[J/mol K]} }{ 28.97 _{[g/mol]} } = 0.287 J/g K \quad or \quad 287 J/kg K }$
Specific Gas Constant (R) for dry air = 287 J/kg K
8.314462 [J/mol K] / 28.97 [g/mol] = 0.287 J/g K
Ideal Gas Law using the Specific Gas Constant
Ideal Gas Law [using the Specific Gas Constant] shows the relationship of Pressure, Volume, and Temperature, within a specific Ideal Gases:
$latex \boxed{ p V = mRT } &s=2$
Properties of dry air using ISA standard conditions at Sea Level:
- p = Absolute pressure (101325 Pa)
- V = Volume (1 m^3)
- m = mass (1.22521 kg)
- R = Specific Gas Constant for dry air (287 J/kg K)
- T = Absolute temperature (288.15 K)
Example: reorder the equation to solve for temperature.
$latex p V = m R T &s=0$ becomes $latex \frac{p V}{m R} = T &s=2$
$latex \boxed{ \frac{ 101325 _{[Pa]} \times 1 _{[m^3]} }{ 1.225 _{[kg]} \times 287 _{[J/kg K]}} = 288.15 K }$
(101325 [Pa] x 1 [m^3]) / (1.225 [kg] x 287 [J/kg K]) = 288.15 [K]
288.15 K is the ISA standard temperature at Sea Level, which brings us directly back to where we started.
Simplifying the Specific Version of the Ideal Gas Law
By combining the mass and volume into a density, we can slightly simplify the specific version using $latex \rho = m / V$:
$latex \frac{ p V }{ V } = \frac { m R T }{ V } &s=1$ becomes
$latex \boxed{ p = \rho R T } &s=2$
Background
The Ideal Gas Law corresponds to Boyle’s Law and Charles’ Law.
Boyles Law – Robert Boyle (1662):
The product of pressure and volume is exactly a constant for an ideal gas.
$latex pV = k&s=1$
Charles’ Law – Jacques Charles (1787):
Charles’s law states that if a given quantity of gas is held at a constant pressure, its volume is directly proportional to the absolute temperature.
$latex \frac{V1}{T1} = \frac{V2}{T2}&s=2$
Avogardo’s Constant and moles – Amedeo Avogadro et.al: (1811-)
Avagadro’s Constant is 6.02214 x 10^23
Moles is simply a unit of quantity just like a dozen indicates twelve, and is based on Avogardo’s Constant, where a mol contains 6.02214×10^23 particles, or molecules.
Initial Ideal Gas Statement – Emile Clapeyron (1834):
Combined Boyle’s Law with Charles’ Law into the first statement of the ideal gas.
$latex pV = \bar{R}T_{c} + 273.15&s=1$
Consider
All these ideas and constants are extremely important and is rooted throughout fluids engineering and physics. Practice them until you can say them backwards and forwards, and can translate from volumes to masses with ease.
Credits:
Brilliant Ideal Gas Law image was furnished by