### Ideal gas equation its derivation

### Study ideal gas law in chemistry and physics

A gas at equilibrium has a definite value of pressure, volume, temperature, and composition. These are called state variables and are determined experimentally.The state of the gas can be defined by these variables. Boyle's in 1662, Charles's in 1787 and Avogadro laws give the birth of an equation of state for

**Ideal gas**.

#### Derivation of Ideal gas law

Boyle's law,

V ∝ 1/P

when n and T are constant of a gas.

Charles law,

V ∝ T

when n and P are constant for a gas.

Avogadro's law,

V ∝ n

when P and T are constant for a gas.

V ∝ 1/P

when n and T are constant of a gas.

Charles law,

V ∝ T

when n and P are constant for a gas.

Avogadro's law,

V ∝ n

when P and T are constant for a gas.

When all the variables are taken into account, the variation rule states that,

V ∝ (1/P) × T × n

or, V = R × (1/P) × T × n

or, V = R × (1/P) × T × n

Ideal gas law |

**ideal gas law**of the state. This equation is found to hold most satisfactory when pressure tense to zero.

At ordinary temperature and pressure, the equation is found to deviated about 5%. Real gases attain ideal behavior only at low pressures and very high temperatures.

#### Value of universal gas constant

At NTP 1 mole gas at 1-atmosphere pressure occupied 22.4 lit of gas.Ideal gas law

PV =RT

or, R = (PV)/(nT).

PV =RT

or, R = (PV)/(nT).

Putting the values of pressure, volume, and temperature in the above equation.

R = ( 1 atm × 22.4 lit)/(1 mole × 273 K)

= 0.082 lit atm mol⁻¹ K⁻¹

Problem= 0.082 lit atm mol⁻¹ K⁻¹

Derive the value of

**gas constant**R when pressure is expressed in the atm, and volume in cm³ and pressure in dyne m⁻² and volume mm³.

Answer

- 82.05 atm cm³ mol⁻¹ K⁻¹
- 8.314 × 10¹⁴ dyne m⁻² mm³ mol⁻¹ K⁻¹

#### Universal gas constant in CGS and SI units

Pressure = 1 atm = 76 cm Hg

= 76 cm × 13.6 gm cm⁻² × 981 cm sec⁻²

= 76 × 13.6 × 981 dyne cm⁻²

Volume = 22.4 × 10³ cm³

Temperature = 273 K

∴ R = (76 × 13.6× 981 × 22.4 × 10³)/(1 × 273)

= 8.314 × 10⁷ dyne cm² mol⁻¹ K⁻¹

Work = force × displacement.

∴ erg = dyne cm².

R = 8.314 × 10⁷ erg mol⁻¹ K⁻¹

1 J = 10⁷ erg.

Thus universal constant of gas in SI units

= 8.314 J mol⁻¹ K⁻¹.

4.18 J = 1 calories,

∴ R = 8.314 / 4.18 calories mol⁻¹ K⁻¹

= 1.987 calories mol⁻¹ K⁻¹

≃ 2 calories mol⁻¹ K⁻¹

Problem= 76 cm × 13.6 gm cm⁻² × 981 cm sec⁻²

= 76 × 13.6 × 981 dyne cm⁻²

Volume = 22.4 × 10³ cm³

Temperature = 273 K

∴ R = (76 × 13.6× 981 × 22.4 × 10³)/(1 × 273)

= 8.314 × 10⁷ dyne cm² mol⁻¹ K⁻¹

Work = force × displacement.

∴ erg = dyne cm².

R = 8.314 × 10⁷ erg mol⁻¹ K⁻¹

1 J = 10⁷ erg.

Thus universal constant of gas in SI units

= 8.314 J mol⁻¹ K⁻¹.

4.18 J = 1 calories,

∴ R = 8.314 / 4.18 calories mol⁻¹ K⁻¹

= 1.987 calories mol⁻¹ K⁻¹

≃ 2 calories mol⁻¹ K⁻¹

What is the value of gas constant R when pressure is expressed in Torr and volume in dm³?

Answer

61.54 Torr dm³ mol⁻¹ K⁻¹

#### What is the significance of gas constant?

For n mole ideal gas

PV = nRT

or, R = PV/nT.

∴ Unit of R = (unit of P × unit of V)/(unit of n × unit of T).

PV = nRT

or, R = PV/nT.

∴ Unit of R = (unit of P × unit of V)/(unit of n × unit of T).

Pressure = (force/area)

= (force/ length²)

= force × length⁻².

Volume = length³.

= (force/ length²)

= force × length⁻².

Volume = length³.

∴ R = (force × length⁻² × length³)/(amount of gas × kelvin)

= (force × length)/(amount of gas × kelvin)

= (Work or Energy)/(amount of gas × kelvin).

= (force × length)/(amount of gas × kelvin)

= (Work or Energy)/(amount of gas × kelvin).

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

#### The molecular weight of ideal gas

n mole ideal gas

PV = nRT

or, PV= (g/M)RT

where g = weight of the gas in gram and M = molecular weight of the gas.

P = ( g/V) (RT/M)

Density = weight of the gas/volume

or, d = g/V

∴ P = dRT/M

From this equation, we can easily find out the molecular weight of a gas.PV = nRT

or, PV= (g/M)RT

where g = weight of the gas in gram and M = molecular weight of the gas.

P = ( g/V) (RT/M)

Density = weight of the gas/volume

or, d = g/V

∴ P = dRT/M

Problem

The volume of 12.8 grams of gas at 760 mm-Hg pressure and 27° C is 10 liter. Calculate the molecular weight of this gas.

Answer

Ideal gas law, PV = nRT.

or, PV = (g/M) × RT

∴ M = gRT/PV

= (12.8 × 0.082 × 300)/(1 × 10).

= 31.49 gm mol⁻¹

Problemor, PV = (g/M) × RT

∴ M = gRT/PV

= (12.8 × 0.082 × 300)/(1 × 10).

= 31.49 gm mol⁻¹

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

The molecular weight of ideal gas,

M = dRT/P

∴ The molecular weight of ammonia

= (3.42 × 0.082 × 303)/5 gm mol⁻¹

= 16.99 gm mol⁻¹

≃ 17 gm mol⁻¹

M = dRT/P

∴ The molecular weight of ammonia

= (3.42 × 0.082 × 303)/5 gm mol⁻¹

= 16.99 gm mol⁻¹

≃ 17 gm mol⁻¹

#### Gas molecules present in an ideal gas

Ideal gas law for n mole gas,

PV = nRT.

or, PV= (N/N₀) RT

N =

N₀ = Avogadro number = 6.023 × 10²³.

∴ P = (N/V) × (R/N₀) × T

∴ P = N′ kT

N′ = gas molecules present per unit volume.

k = Boltzmann constant = R/N₀

= 1.38 × 10⁻¹⁶ erg molecules⁻¹ K⁻¹

ProblemPV = nRT.

or, PV= (N/N₀) RT

N =

**gas molecules**present in an ideal gas,N₀ = Avogadro number = 6.023 × 10²³.

∴ P = (N/V) × (R/N₀) × T

∴ P = N′ kT

N′ = gas molecules present per unit volume.

k = Boltzmann constant = R/N₀

= 1.38 × 10⁻¹⁶ erg molecules⁻¹ K⁻¹

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

= (7.6 × 10⁻³ mm Hg) (101.235 kPa/760 mm Hg)

= 1.01235 × 10⁻³ kPa.

Ideal gas law for n moles gas

PV = nRT

or, n = PV/RT

= (1.01235 × 10⁻³ × 10⁻⁶)/(8.314 × 273)

= 4.46 × 10⁻¹³ mole

Number of gaseous molecules

N = n N₀.

= (4.46 × 10⁻¹³ mole)(6.023 × 10⁻²³ mol⁻¹)

= 2.68 × 10⁻¹¹.

ProblemPressure (P) = 7.6 × 10⁻³ mm Hg

= (7.6 × 10⁻³ mm Hg) (101.235 kPa/760 mm Hg)

= 1.01235 × 10⁻³ kPa.

Ideal gas law for n moles gas

PV = nRT

or, n = PV/RT

= (1.01235 × 10⁻³ × 10⁻⁶)/(8.314 × 273)

= 4.46 × 10⁻¹³ mole

Number of gaseous molecules

N = n N₀.

= (4.46 × 10⁻¹³ mole)(6.023 × 10⁻²³ mol⁻¹)

= 2.68 × 10⁻¹¹.

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

Answer

Moles of ethane

n₁ = 2/30 = 0.0667

Moles of carbon dioxide

n₂ = 3/44 = 0.0682

Total moles (n₁ + n₂) = (0.0667 + 0.0682)

= 0.1349

∴ Total pressure (P) = (n₁ + n₂)RT/V

or, P = (0.1349 × 0.082 × 323)/5

= 0.715 atm

Problemn₁ = 2/30 = 0.0667

Moles of carbon dioxide

n₂ = 3/44 = 0.0682

Total moles (n₁ + n₂) = (0.0667 + 0.0682)

= 0.1349

∴ Total pressure (P) = (n₁ + n₂)RT/V

or, P = (0.1349 × 0.082 × 323)/5

= 0.715 atm

At 2 atm constant pressure slope of a one-mole ideal gas in V vs T graph is x L mol⁻¹ K⁻¹. Find out the value of gas constant R by x.

Answer

Ideal gas law for 1-mole gas,

PV = nRT

or, V = (nR/P) × T.

The slope of the V vs T graph = nR/P

where P = 2 atm.

∴ (R/2 atm) = x L mol⁻¹ K⁻¹

Value of gas constant R = 2x L atm mol⁻¹ K⁻¹

PV = nRT

or, V = (nR/P) × T.

The slope of the V vs T graph = nR/P

where P = 2 atm.

∴ (R/2 atm) = x L mol⁻¹ K⁻¹

Value of gas constant R = 2x L atm mol⁻¹ K⁻¹