Ideal gas law formula derivation

Ideal gas law equation of state

An ideal gas law formula represents the relationship between pressure, volume, the temperature of a given mass of a gas. Thus at equilibrium, gas has a definite value of pressure, volume, temperature, and composition. These are called state variables and this equation is known as the equation of state.

Ideal gas law formula derivation
Volume change of the gas at constant T

Boyle’s in 1662, Charles’s in 1787 and Avogadro give the birth of the ideal gas law equation state.

Derivation of the ideal gas law

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.

Thus when all the variables are taken into account, the variation rule states as,

V \propto\frac{1}{P}\times T\times n

or,V =R\times \frac{1}{P}\times T\times n

PV = nRT
where R = universal gas constant.

Hence this equation is given the relation between pressure, volume, temperature, and composition. 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 gases attain ideal behavior only at low pressures and very high temperatures.

Universal gas constant in different units

Universal gas constant R
atm unit 0.082 lit atm mol-1 K-1
CGS unit 8.314×107 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 gas at 1-atmosphere pressure occupied 22.4 lit of volume. Thus from the ideal gas law

R=\frac{PV}{nT}=\frac{1atm\times 22.4lit}{1mol\times 273K}

=0.082\, lit\, atm\, mol^{-1}K^{-1}

Universal gas constant in CGS units

In SI units, pressure = 1 atm
= 76 × 13.6 × 981 dyne cm-2

\therefore R=\frac{76\times 13.6\times 981\times 22.4\times 10^{3}}{1\times 273}

= 8.314 × 107 dyne cm2 mol-1 K-1
∴ R = 8.314 × 107 erg mol-1 K-1
where dyne cm2 = erg

Universal gas constant in SI units

R = 8.314 × 107 erg mol-1 K-1
But 1 J = 107 erg

Thus universal constant of gas in SI units
= 8.314 J mol-1 K-1

Agin 4.18 J = 1 calorie

\therefore R=\frac{8.314}{4.18}\, calories\, mol^{-1}\, K^{-1}

= 1.987 calories mol-1 K-1
≃ 2 calories mol-1 K-1

Significance of ideal gas constant

For n mole ideal gas
PV = nRT

\therefore unit\, of\, R= \frac{unit\, of\, P\times unit\, of\, V}{unit\, of\, n\times unit\, of\, T}

But pressure = force/area = force/ length2
= force × length-2

Volume = length3

\therefore R=\frac{force\times length^{-2}\times length^{3}}{amount\, of\, gas\times kelvin}

=\frac{work\, or\, energy}{amount\, of\, gas\times kelvin}

Thus universal gas constant R = energy per mole per kelvin or amount of work or energy that can be obtained from one mole of gas when its temperature raised by one kelvin.

Density from the ideal gas equation

The ideal gas equation for n mole

PV=nRT=\left ( \frac{g}{M} \right )RT

where g = weight in gram
M = molecular weight

\therefore P= \left ( \frac{g}{V} \right ) \left ( \frac{RT}{M} \right )= \frac{dRT}{M}

where d = density = weight/volume

Thus from the above equation, we can easily find out the density from the known molecular weight of the gas.

Problem
The density of ammonia at 5-atmosphere pressure and 30°C temperature 3.42 gm lit-1. What is the molecular weight of ammonia?

Answer

Molecular\, weight(M)=\frac{dRT}{P}

\therefore M_{NH_{3}}=\frac{3.42\times 0.082\times 303}{5}

= 16.99 gm mol-1
≃ 17 gm mol-1

Number of gas molecules per unit volume

PV=nRT=\left (\frac{N}{N_{0}} \right )RT

where N = number gas molecules present
N0 = Avogadro number = 6.023 × 1023

\therefore P=\left ( \frac{N}{V} \right )\times \left ( \frac{R}{N_{0}} \right )\times T={N}'kTwhere N′ = gas 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 dm3

Pressure (P) = 7.6 × 10-3 mm Hg

=\frac{7.6\times 10^{-3}mmHg\times 101.235kPa}{760mmHg}

= 1.01235 × 10-3 kPa

\therefore {N}'=\frac{1.01235\times 10^{-3}}{1.38\times 10^{-9}\times 273}

= 2.68 × 10-11

The pressure of an ideal gas derivation

Assuming ideal behavior finds out the total pressure exerted by 2 gm methane and 3 gm carbon dioxide contained in a vessel of 5-liter capacity at 50°C.

Moles\, of\, CH_{4}\left ( n_{1} \right )=\frac{2}{16}=0.125

Moles\, of\, CO_{2}\left ( n_{2} \right )=\frac{3}{44}=0.0682

Total moles (n1 + n2) = (0.125 + 0.0682)
= 0.1932

\therefore P_{total}=\frac{0.1932\times 0.082\times 323}{5}=5.30\, atm

Thus the total pressure of this ideal gas mixture = 5.30 atm.