## Ideal Gas Law in Physics Chemistry

**Ideal gas law** represents the mixed relationship between pressure, volume, the temperature of gases for study properties of gases in physics and chemistry. This equation connecting these state variables by the universal gas constant (R) of the ideal gas. Therefore the ideal or perfect gas law formula can use for calculating the value of pressure, volume, temperature, and the number of gas molecules per unit density.

Boyle’s in 1662, Charles’s in 1787, and Avogadro give the derivation formula of the ideal gas or perfect gases.

### Formula of the ideal gas equation of state

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, this equation 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%. Thus real gas attain perfect ideal behavior only at low pressures and very high temperatures.

### Universal gas constant value

Value of Universal gas constant R | |

atm unit | 0.082 lit atm mol^{-1} K^{-1} |

^{CGS unit} |
8.314×10^{7} erg mol^{-1} K^{-1} |

SI unit | 8.314 joule mol^{-1} K^{-1} |

Cal unit | 2 cal mol^{-1} K^{-1} |

At NTP 1 mole gases at 1-atmosphere pressure occupied 22.4 lit of volume. Thus 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 SI units, pressure = 1 atm

= 76 × 13.6 × 981 dyne cm^{-2}

∴ Universal gas constant values in CGS units

= (7.6×13.6×981×22.4×10^{3})/(1×273)

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

∴ R = 8.314 × 10^{7} erg mol^{-1} K^{-1}

where dyne cm^{2} = erg

#### Value of universal gas constant in SI unit

R = 8.314 × 10^{7} erg mol^{-1} K^{-1}

But 1 J = 10^{7} erg

Thus universal constant of gas in SI units

= 8.314 J mol^{-1} K^{-1}

Agin from heat relation

4.18 J = 1 calorie

∴ R = (8.314/ 4.18) cal mol^{-1} K^{-1}

= 1.987 calories mol^{-1} K^{-1}

≃ 2 calories mol^{-1} K^{-1}

### Significance of Ideal Gas Law

For n mole ideal gases

PV = nRT

Therefore, from the significance of the ideal gas law, 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 equation of state for n mole

PV = nRT = (g/M) × RT

where g = weight in gram

M = molar mass

∴ P = dRT/M

where d = density = g/V

Therefore, from the ideal gas equation formula, we can easily find out the density 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}. What is the molar mass of ammonia?

Answer

Molar mass (M) = dRT/P

∴ 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 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.

Answer

Volume (V) = 1 ml = 10^{-6} dm^{3}

Pressure (P) = 7.6 × 10^{-3} mm Hg

= 1.01235 × 10^{-3} kPa

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

= 2.68 × 10^{-11}

### Pressure of the Mixed Gases

Assuming perfect behavior finds out the mixed pressure exerted by 2 gm 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 this mixed gas molecules = 5.30 atm.