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Definition of Ideal Gas in simple terms is “A theoretical gas that constitutes a random combination of point particles and elastic collision is the method of interaction between these particles”. An ideal gas is completely different when compared with real gas in many aspects. We have to remember that ideal gas responds perfectly to all the equations of ideal gas law, so the ideal gas formula is very important.
You can easily identify an ideal gas if all the particles are interacting perfectly as elastic or point-like collision. Sometimes the pressure and temperature tend to change which leads to many gases like nitrogen, oxygen, hydrogen, and even carbon dioxide behaving like an ideal gas. This scenario can be witnessed when the temperature is high and the pressure is low. When compared with the particle’s kinetic energy, the potential energy will be significantly low. In this article, the student will learn precisely what is an ideal gas law statement, derivation, application, volume of an ideal gas at STP & NTP, values of gas constant (R) in different Units, and a couple of solving equations.
Ideal Gas Law statement
If you take the volume of 1gram molecules from an ideal gas and product of pressure, then we can derive the Universal gas and absolute gas temperature. In the year 1834, Benoit Paul Emile Clapeyron defined ideal gas law as an amalgamation of Boyle’s law, Gay-Lussac’s law, Charles’s law, and Avogadro’s law
If you hold a mass of gas under persistent temperature then you will notice that gas pressure will be exactly opposite to the volume of Gas. So, the gas volume and pressures product have a sustaining connection with the temperature and the ideal gas constant
Ideal Gas Law
PV = nRT
- P = Pressure of the ideal gas
- V = Volume of the ideal gas
- n = Amount of substance of Ideal Gas (Measures in moles)
- R = Ideal gas constant
- T = Temperature
Based on S.I. systems, the ideal gas constant value is 8.314 J mol-1K-1
Ideal Gas Law derivation
If we observe the works of Gay-Lussac, Robert Boyle, and Amedeo Avogadro we can easily derive the ideal gas law. The experts have derived their own observations, so we have to combine their observations to derive a single equation that will show us that volume is equivalent to the total number of moles and it’s the exact opposite in terms of temperature and pressure.
| Scientist | Ideal Gas Law derivation |
|---|---|
| Charles Law | V ∝ T |
| Boyle’s Law | V ∝ 1/P |
| Avogadro’s Law | V ∝ n |
| Combined ideal gas law derivation = V ∝ nT/P | |
As per Boyle’s Law, the value of n and value of T is persistent, then the gas exerts a pressure that has inversely proportional volume.
As per Charles’s Law, the value of p and value of n is persistent, then the temperature can be seen as directly proportional to the volume of gas.
As per Avogadro’s Law, the value of P and value of T is persistent, then the total number of moles of the gas can be seen as directly proportional to the volume of the gas.
So, let’s write all the expressions as,
V = RnT/P = nRT/P
(Now let us multiply both the sides by P, we get)
PV = nRT
Hence, the Ideal Gas Law derivation is explained in detail.
Ideal Gas Law applications
The basic applications of ideal gas law are density, molar mass, and volume. According to Ideal Gas Law PV = nRT, if there is any change in temperature and pressure, there will be variation in the number of moles in a given volume of gas and volume of a given quantity of gas. The STP (Standard temperature & Pressure) of an ideal gas is 273.15K & 1 atm has a volume of 22.4L (standard molar volume).
- We have already derived the universal ideal gas law which comprises volume, pressure, temperature, number of moles of gas with whatever the gas chemical identity is.
PV = nRT
- Let us take the density of a gas ‘d’. Density is nothing but molar mass ratio over volume. The equation derived is,
d=m/v
- The definition of Molar mass is termed as mass over moles, that is ration m/n,
M=m/n
- Then let’s derive an equation for density as,
d = MP/RT
The density of a gas is directly proportional to molar mass and pressure but in terms of temperature, it is inversely proportional. Some of the ideal gas law application examples are,
- Carbondioxide (CO2) is heavier than Nitrogen (N2) because the molar mass of CO2 is 44 g/mol and N2 is 28 g/mol. It is denser than air. When we release CO2 from a fire extinguisher, it engulfs the fire and prevents O2 from reaching the inflammable.
- During high temperatures, the gases with the same molar masses tend to have a density level very low. This is the reason which supports the lifting of hot air balloons to float.
Volume of an Ideal Gas at STP
The volume of an ideal gas at STP is 22.41 L/mol. Let us see how this volume is derived. Let us take 1 mole of an ideal gas, that is STP = 0oC, and 1 atm. The temperature is 273K
Let us take Ideal Gas Equation,
PV = nRT
1V = 1(0.082)(273.15)
V = 22.41 L
The volume of an Ideal Gas at STP (standard temperature and pressure)
is 22.41 L/Mol
Volume of an Ideal Gas at NTP
The volume of an ideal gas at NTP is 24.04L. Let us see how this volume is derived. Let us take 1 mole of an ideal gas, pressure is 1atm, temperature is 293K or 20C.
Let us take Ideal Gas Equation,
PV = nRT
1V = 1(293)(0.082)
V = 24.04L
The volume of an Ideal Gas at NTP (Normal temperature and pressure)
is 24.04 L/Mol
Values of Gas Constant (R) in different Units (Tabular form)
The value ‘R’ in the ideal gas law is known as the gas constant. The value of R will keep changing based on volume and different units of pressure. When we use the Ideal Gas Equation, the temperature will be mentioned in Kelvin only. You must know the exact value of R to derive the correct solution for the question. Some of the Values of gas constant R in different units are mentioned below,
| Values of Gas Constant (R) | Different Units |
|---|---|
| 8.314 4621 | J K−1 mol−1 |
| 0.082 057 | L atm K−1 mol−1 |
| 62.363 | L Torr K−1 mol−1 |
| 8.314 | J K−1 mol−1 |
| 8.314 | m3 Pa K−1 mol−1 |
| 62.363 | L mmHg K−1 mol−1 |
Solved examples based on Derivation of Ideal Gas Equation
Ques. From the given details, find the volume of an ideal gas? Neon (5g) with pressure 256 mm hg. The temperature is 35C.
Ans. First, identify all the given information. P is 256 mm hg, temperature (T) is 35C, R is 0.082057 L atm K−1 mol−1, mass (m) is 5g, and we have to find the V,
Now,
P = 256 x 1 / 760 = 0.3368
M = 5 x 1 / 20.1797 = 0.25
T = 35C + 273 = 308K
V = nRT / P
V = (0.25)(0.08206)(308) / 0.3368
V = 19L
Ques. From the given details, find the pressure exerted by the gas? 1 mole of an ideal gas, Volume 1 cubic meter, the temperature is 300 K.
Ans. First, identify all the given information. V is 1, n is 1, T is 300.
PV = nRT
P x 1 = 1 x 25/3 x 300
P = 2500 Pa
Ques. Find out the Volume occupied by carbon dioxide (2.34 g) at Standard Temperature & Pressure (STP)?
Ans. The basic Ideal gas law is PV = nRT. So first we have to re arrange the formula as we have to find the Volume. So, the rearrangement will be as follows,
PV = nRT
V = nRT / P
V = (2.34 / 44.0) (0.08206) (273.0) / 1
V = 1.19 L
Ques. From the given details, find the temperature of neon gas. Moles of neon gas are 0.654 and occupy 12.30 L at 1.95 pressure.
Ans. First, identify all the given information.
P = 1.95 atm
V = 12.30 L
n = 0.654
R = 0.08206 L atm K−1 mol−1
T = ?
The basic Ideal gas law is PV = nRT. So first we have to re-arrange the formula as we have to find the Temperature. So, the rearrangement will be as follows,
PV = nRT
T = PV/nR
T = (1.95)(12.30) / (0.654)(0.08206)
T = 447K
Ques. From the given details, find the Molecular weight of the gas. A gas (96.0g) occupies 48L at a pressure of 700mmhg, temperature is 20C.
Ans. The basic Ideal gas law is PV = nRT. So first we have to re-arrange the formula as we have to find the molecular weight. So, the rearrangement will be as follows,
PV = nRT
n = PV / RT
n = (700/760)(48) / (0.08206)(293.0)
n = 1.8388 mol
Now we have to divide the derived mole with the total grams given, (1.8388 / 96 = 52.2). So, the molecular weight of the given gas is 52.2 g/mol.
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