## What is ideal gas law?

**Ideal gas law** or **perfect gas law** represents the mixed relationship between pressure, volume, and temperature of gases for learning the physical properties of the gas molecule in physics or chemistry. The **ideal gas equation** balances these state variables in terms of the universal gas constant (R).

The ideal or perfect gas law formula can use for calculating the value of pressure, volume, temperature, diffusion or effusion, concentration, and the number of gas molecules per unit volume or density of a gas.

Boyle’s in 1662, Charles’s in 1787, and Avogadro laws are used to derive ideal or perfect gas equation or formula. The kinetic theory of gases derives properties of ideal gases.

Four thermodynamics variables (pressure, volume, temperature, and mole number) are used to derive the ideal gas equation from various gas laws. Some of these depend on the mass of the system while others are independent of the mass.

- In thermodynamics, the property which proportional to the mass of the system is called intensive property.
- The property of the system which is independent of the mass of the system is called intensive property.

In ideal gas law, the volume is an intensive property but temperature and pressure are extensive properties.

## Ideal gas law formula derivation

Boyle’s, Charles’s, and Avogadro’s laws give the general derivation formula of the ideal or perfect gases. Mathematical derivation of these laws are,

- Boyles law V ∝ 1/T, when n and T are constant.
- Charles law, V ∝ T, when n and P are constant.
- Avogadro’s law, V ∝ n, when P and T are constant.

When all the variables of gas laws are taken into account, we find out the mathematical expression of the ideal gas law equation,

**PV = nRT**

where R = **universal gas constant**.

Ideal gas law defines the relation between pressure, volume, temperature, and composition of gases. But the equation is found to hold most satisfaction when pressure is low or tense to zero.

At ordinary temperature and pressure, the equation is found to deviate about 5%. Therefore, the real or Van der Waals gas obeys the ideal gas law only at low pressures and very high temperatures.

## Universal gas constant value

Universal constant values and unit dimension can be calculated from the ideal gas law,

PV = nRT

At NTP 1 mole gases at 1 atmosphere pressure occupied 22.4 lit of volume.

From the ideal gas equation,

R = PV/nT

= (1 atm × 22.4 lit)/(1 mol × 273 K)

= 0.082 lit atm mol^{−1} K^{−1}

### Universal gas constant in CGS unit

In CGS units, pressure = 1 atm = 76 × 13.6 × 981 dyne cm^{−2} and volume = 22.4 liter = 22.4 × 10^{3} cm^{3}.

Putting the values of P, V, T, and n in ideal gas law, we have universal gas constant,

R = (7.6 × 13.6 × 981 × 22.4 × 10^{3})/(1 × 273)

= 8.314 × 10^{7} dyne cm^{2} mol^{−1} K^{−1}

= 8.314 × 10^{7} erg mol^{−1} K^{−1}

Here, dyne cm^{2} = erg

### Universal gas constant in SI unit

The values of universal constant (R) in the CGS system = 8.314 × 10^{7} erg mol^{−1} K^{−1}. But 1 J = 10^{7} erg.

Therefore, the universal constant in the SI unit,

= 8.314 J mol^{−1} K^{−1}

Agin from specific heat relation, 4.18 J = 1 calorie. Therefore, the universal gas constant,

R = (8.314/ 4.18) cal mol^{−1} K^{−1}

= 1.987 calories mol^{−1} K^{−1}

≃ 2 calories mol^{−1} K^{−1}

From the ideal gas equation,

R = PV/nT

= (1 bar × 22.711 dm^{3})/(1mol × 273.15 K)

= 0.08314 bar dm^{3} mol^{−1} K^{−1}

It is another SI unit of universal gas constant (R).

**Problem**: Determine the value of gas constant R when pressure is expressed in Torr and volume in dm^{3}.

**Solution**: By definition, 1.01325 bar = 760 Torr.

Hence, R = (0.08314 × 760)/1.01325 Torr dm^{3} mol^{−1} K^{−1}

= 62.36 Torr dm^{3} mol^{−1} K^{−1}

## R in ideal gas law

For n mole ideal gases, PV = nRT or R = PV/nT. The unit of universal gas constant = (pressure × volume)/(amount of gas × temperature).

The unit of pressure = force length^{−2} and volume = length^{3}.

Thus the unit of R = (force × length^{−2} × length^{3})/(amount of gas × temperature) = (force × length)/(amount of gas × temperature)

From the definition, work or energy = force × length

Therefore, the unit of R = (work or energy)/(amount of gas × temperature).

In learning chemistry or physics, the dimensions R obtained from the ideal gas law is energy or work per mole per kelvin. Therefore, R represents the amount of work or energy that can be obtained from one mole of gas when its temperature is raised by one kelvin.

## Density from ideal gas law

The ideal gas law for n mole,

PV = nRT = (g/M) × RT

where g = weight in gram, M = molar mass

Again, g/V = d = density

Thus, P = dRT/M

Therefore, the ideal gas law formula is used to find out the density of gases from the known molar mass of gases.

**Problem:** The density of ammonia at 5 atmosphere pressure and 30°C temperature is 3.42 g lit^{−1}. How can we calculate the molar mass of ammonia from the ideal gas equation?

**Solution:** From the density formula,

molar mass (M) = dRT/P

Therefore, the molecular mass of ammonia,

M_{NH3} =(3.42 × 0.082 × 303)/5

= 16.99 g mol^{−1}

≃ 17 g mol^{−1}

## Number of molecules per unit volume

PV = nRT = (N/N_{0}) × RT

where N = number gas molecules present

N_{0} = Avogadro number = 6.023 × 10^{23}

∴ P = (N/V) × (R/N_{0}) × T = N′KT

where N′ = number molecules per unit volume

k = Boltzmann constant = R/N_{0}

= 1.38 × 10^{−16} erg molecules^{−1} K^{−1}

**Problem: **Using the ideal gas equation, estimate the number of gaseous molecules left in a volume of 1 ml if it is pumped out to give a vacuum of 7.6 × 10^{−3} mm of Hg at 0°C.

**Solution:** Volume (V) = 1 ml = 10^{−6} dm^{3}

Pressure (P) = 7.6 × 10^{−3} mm Hg

= 1.01235 × 10^{-3} kPa.

Therefore, the number of gas molecules,

N’ = (1.01235 × 10^{−3})/(1.38 × 10^{−9} × 273)

= 2.68 × 10^{−11}

## How to find total pressure of gas mixture?

The ideal gas equation is used to find out the total pressure of the gas mixture. If a mixture contains n_{1} mole of gas A and n_{2} mole of gas b, the total pressure of the gas mixture = (n_{1} + n_{2})RT.

**Problem**: At 50 °C, an ideal gas mixture containing 2 g methane gas (organic compound) and 3 g carbon dioxide gas molecules in a vessel of 5 liter capacity. Find out the total pressure of the gas mixture.

**Solution**: From the above problem,

Mole number of methane (n_{CH4}) = 2/16 = 0.125

Mole number of carbon dioxide (n_{CO2}) = 3/44 = 0.0682

Hence total moles, n_{total} = (n_{CH4} + n_{CO2}) = (0.125 + 0.0682)= 0.1932.

Therefore, the total pressure of gas mixture from ideal gas law,

Ptotal = (n_{total} × RT)/V

= (0.1932 × 0.082 × 323)/5

= 5.30 atm

The total pressure of a gas mixture derived from the ideal gas law or equation is used for the calculation of the mixed pressure for different gases like oxygen, nitrogen, carbon dioxide, hydrocarbon, etc.