## Ideal Gas Law Formula

**Ideal gas law** or **perfect gas law** represents the mixed relationship between pressure, volume, the temperature of gases for learning the physical properties of the gas molecule in physics or chemistry. Therefore, the ideal gas equation balancing these state variables in terms of 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. Boyle’s in 1662, Charles’s in 1787, and Avogadro give the derivation formula of the ideal or perfect gases, and the kinetic theory of gas provides the properties of ideal gases.

There are four variables in the perfect or ideal gas law of states, P, V, T, and n. 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. Property of the system which independent of the mass of the system is called intensive property. Therefore volume in the ideal gas equation is an intensive property but temperature, pressure in ideal gas law are extensive properties.

### Ideal gas Law Formula Derivation

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 are taken into account, the variation rule states as,

V ∝ (T × n)/P

or, V = (R × T × n)/P

PV = nRT

where R = universal gas constant.

Therefore, ideal gas law is given the relation between pressure, volume, temperature, and composition of gases. But the equation found to hold most satisfactory when pressure low or tense to zero. At ordinary temperature and pressure, the equation found to deviated about 5%. Therefore, the real gas or Van der Waals gas attains perfect ideal behavior only at low pressures and very high temperatures.

### Universal Gas Constant Value

Universal constant values unit and dimension can be calculated from ideal gas law, PV = nRT. At NTP 1 mole gases at 1-atmosphere pressure occupied 22.4 lit of volume. Therefore, 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}.

#### Value of 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 × 103 cm^{3}. Therefore, 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}, where dyne cm^{2} = erg.

#### Value of Universal Gas Constant in SI-unit

The Values of universal constant (R) in CGS-system = 8.314 × 10^{7} erg mol^{-1} K^{-1}. But 1 J = 10^{7} erg. Therefore, the universal constant in SI units = 8.314 J mol^{-1} K^{-1}. Agin from specific heat relation, 4.18 J = 1 calorie. Therefore universal gas constant from ideal gas law equation = (8.314/ 4.18) cal mol^{-1} K^{-1} = 1.987 calories mo^{l-1} K^{-1} ≃ 2 calories mol^{-1} K^{-1}.

### Significance of Ideal Gas Law

For n mole ideal gases, PV = nRT or R = PV/nT. Therefore, the unit of universal gas constant = (unit of pressure × unit of volume)/(amount of gas molecule × unit of temperature). Here, the unit of pressure = force length^{-2} and volume = length^{3}. Therefore, the unit of R = (force × length)/(amount of gas molecule × unit of temperature), where force × length = work or energy.

Therefore, from the significance of the ideal gas law in learning chemistry or physics, R = energy per mole per kelvin or amount of work or energy that can be obtained from one mole of gases when its temperature raised by one kelvin.

### Formula of Ideal Gas Density

The ideal gas law for n mole, PV = nRT = (g/M) × RT, where g = weight in gram, M = molar mass. Therefore, P = dRT/M, where d = density = g/V. Therefore, from the ideal gas law formula, we can easily find out the density of gaseous chemical elements in chemical science from the known molar mass of the mixed gases.

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

Answer: Molar mass (M) = dRT/P Therefore, molecular mass of ammonia, M_{NH3} =(3.42 × 0.082 × 303)/5 = 16.99 gm mol-1≃ 17 gm 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/N₀

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

Problem: Estimate from the ideal gas law, the number of gaseous molecules left in a volume of 1 mi-liter if it 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}.

### Pressure of the Mixed Gases

Assuming perfect behavior or obeying ideal gas law to finds out the mixed pressure exerted by 2 gm organic hydrocarbon like methane and 3 gm carbon dioxide gas molecules in a vessel of 5-liter capacity at 50°C.

n_{CH4} = 2/16 = 0.125

n_{CO2} = 3/44 = 0.0682

Total moles (n_{1} + n_{2}) = (0.125 + 0.0682)

= 0.1932

∴ P_{total} = (0.1932 × 0.082 × 323)/5 atm

= 5.30 atm

Therefore, the total pressure of mixed ideal gas molecules can be derived from ideal gas law and uses for calculation of the mixed pressure for different gases like oxygen, nitrogen, carbon dioxide, hydrocarbon, etc.