### Ideal gas equation its derivation

- A gas at equilibrium has definite value of pressure(P), volume(V), temperature(T) and composition(n). These are called state variables and are determined experimentally.

- The state of the gas can be defined by these variables. Boyle's(1662), Charles's(1787) and Avogadro laws give the birth of an equation of state for Ideal gas.

### Ideal Gas Equation its derivation

Boyle's law,

V ∝ 1/P

when n and T are constant for a gas

Charl's 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

V ∝ 1/P

when n and T are constant for a gas

Charl's 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

Ideal gas equation |

- Where R is the universal gas constant. This is called the ideal gas equation of state for an ideal gas.

- This equation is found to hold most satisfactory when P tense to zero. At ordinary temperature and pressure, the equation is found to deviated about 5%.

#### Value of universal gas constant

- At NTP 1 mole gas at 1 atm Pressure occupied 22.4 lit of gas.

Thus, R = (PV)/(nT)

Putting the values above equation,

We have, R = ( 1 atm × 22.4 lit)/(1 mol × 273 K)

= 0.082 lit atm mol⁻¹ K⁻¹

Putting the values above equation,

We have, R = ( 1 atm × 22.4 lit)/(1 mol × 273 K)

= 0.082 lit atm mol⁻¹ K⁻¹

#### Value of R in C.G.S. and S.I. system

P = 1 atm = 76 cm Hg

= 76 cm × 13.6 gm cm⁻² × 981 cm sec⁻²

= 76 × 13.6 × 981 dyne cm⁻²

Thus, R = (76 × 13.6× 981 dyne cm⁻² ×22.4 × 10³ cm³)/(1 mol × 273 K)

= 8.314 × 10⁷ dyne cm² mol⁻¹ K⁻¹

Again, Work (W) = Force(F) × Displacement(d),

So, erg = dyne cm².

Thus, R = 8.314 × 10⁷ erg mol⁻¹ K⁻¹

∴ P = dRT/M

= 76 cm × 13.6 gm cm⁻² × 981 cm sec⁻²

= 76 × 13.6 × 981 dyne cm⁻²

Thus, R = (76 × 13.6× 981 dyne cm⁻² ×22.4 × 10³ cm³)/(1 mol × 273 K)

= 8.314 × 10⁷ dyne cm² mol⁻¹ K⁻¹

Again, Work (W) = Force(F) × Displacement(d),

So, erg = dyne cm².

Thus, R = 8.314 × 10⁷ erg mol⁻¹ K⁻¹

We Know That 1 J = 10⁷ erg,

Thus the value of R in S.I. Unit,

= 8.314 J mol⁻¹ K⁻¹

Again, 4.18 J = 1 Cal,

hence, R = 8.314 / 4.18 Cal mol⁻¹ K⁻¹

= 1.987 Cal mol⁻¹ K⁻¹

≃ 2 Cal mol⁻¹ K⁻¹

Thus the value of R in S.I. Unit,

= 8.314 J mol⁻¹ K⁻¹

Again, 4.18 J = 1 Cal,

hence, R = 8.314 / 4.18 Cal mol⁻¹ K⁻¹

= 1.987 Cal mol⁻¹ K⁻¹

≃ 2 Cal mol⁻¹ K⁻¹

### Significance of universal gas constant

The universal gas constant

R = PV/nT

R = PV/nT

- Thus, it has the units of (Pressure × Volume)/(amount of gas × temperature). Now the dimension of pressure and volume are,

Pressure = (force/area)

= (force/ length²)

= force × length⁻² and Volume

= length³

= (force/ length²)

= force × length⁻² and 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)

- Thus, the dimensions of R are energy per mole per kelvin and hence it represents the amount of work or energy) that can be obtained from one mole of gas when its temperature is raised by one kelvin.

### Molar mass from the ideal gas equation

The ideal gas equation is,

PV = nRT

or, PV= (g/M)RT

where g = weight of the gas in gm and M = Molar mass of the gas.

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

We know that, Density (d) = Weight (g)/Volume (V).

PV = nRT

or, PV= (g/M)RT

where g = weight of the gas in gm and M = Molar mass of the gas.

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

We know that, Density (d) = Weight (g)/Volume (V).

∴ P = dRT/M

### Gas molecule present in an ideal gas

The ideal gas equation for n mole gas is,

PV = nRT

Again, PV= (N/N₀) RT

where N = number of molecules present in the gas and N₀ = Avogadro number.

Thus, P = (N/V) × (R/N₀) × T

∴ P = N′ kT

where N′ = number of molecules present per unit volume and

k = Boltzmann constant = R/N₀ = 1.38 × 10-16 erg molecule⁻¹ K⁻¹

PV = nRT

Again, PV= (N/N₀) RT

where N = number of molecules present in the gas and N₀ = Avogadro number.

Thus, P = (N/V) × (R/N₀) × T

∴ P = N′ kT

where N′ = number of molecules present per unit volume and

k = Boltzmann constant = R/N₀ = 1.38 × 10-16 erg molecule⁻¹ K⁻¹

### Questions and answers

Problem- Determine the value of gas constant R when pressure is expressed in Torr and Volume in dm³.

- 61.54 Torr dm³ mol⁻¹ K⁻¹

- Derive the value of R when, (a) pressure is expressed in the atom, and volume in cm³and (b) Pressure in dyne m⁻² and volume mm³.

- (a) 82.05 atm cm³ mol⁻¹K⁻¹

- (b) 8.314 × 10¹⁴ dyne m⁻² mm³ mol⁻¹ K⁻¹

- Find the Molar mass of ammonia at 5 atm pressure and 30⁰C temperature (Density of ammonia = 3.42 gm lit⁻¹).

- 17 gm mol⁻¹

- What is the molecular weight of a gas, 12.8 gms of which occupy 10 liters at a pressure of 750 mm and at 27° C?

We have, PV = nRT

or, PV = (g/M)RT (molecular wt. of N₂ = 28)

∴ P = (7/28) × (0.082 × 300)/3

= 2.05 atm.

Here R = 0.082 lit atm mol⁻¹ K⁻¹

Problemor, PV = (g/M)RT (molecular wt. of N₂ = 28)

∴ P = (7/28) × (0.082 × 300)/3

= 2.05 atm.

Here R = 0.082 lit atm mol⁻¹ K⁻¹

- Calculate the number of molecules present per ml of an ideal gas maintained at a pressure of 7.6 × 10⁻³ mm of Hg at 0°C.

We have given that, V = 1ml = 10⁻⁶ dm³

P = 7.6 × 10⁻³ mmHg

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

= 1.01235 × 10⁻³ kPa

Amount of the gas, n = PV/RT

= (1.01235 × 10⁻³ kPa)(10⁻⁶ dm³)/(8.314 kPa dm³ mol⁻¹ K⁻¹)(273 K)

= 4.46 × 10⁻¹³ mol

Hence the number of molecules, N = n N₀

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

= 2.68 × 10⁻¹¹

P = 7.6 × 10⁻³ mmHg

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

= 1.01235 × 10⁻³ kPa

Amount of the gas, n = PV/RT

= (1.01235 × 10⁻³ kPa)(10⁻⁶ dm³)/(8.314 kPa dm³ mol⁻¹ K⁻¹)(273 K)

= 4.46 × 10⁻¹³ mol

Hence the number of molecules, N = n N₀

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

= 2.68 × 10⁻¹¹

Problem

- Assuming ideal behavior finds out the total pressure exerted by 2 gm ethane and 3 gm CO₂ contained in a vessel of 5-liter capacity at 50°C.

No of molecules of ethane(n₁) = 2/30

= 0.0667

No of molecules of CO₂(n₂) = 3/44

= 0.0682

So the total moles (n₁ + n₂) = (0.0667 + 0.0682)

= 0.1349

Thus the total pressure P = (n₁ + n₂)RT/V

or, P = {0.1349 × 0.082 × (273+50)}/5

= 0.715 atm

= 0.0667

No of molecules of CO₂(n₂) = 3/44

= 0.0682

So the total moles (n₁ + n₂) = (0.0667 + 0.0682)

= 0.1349

Thus the total pressure P = (n₁ + n₂)RT/V

or, P = {0.1349 × 0.082 × (273+50)}/5

= 0.715 atm

Problem

- 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 R by x.

The ideal gas equation for 1-mole gas is,

PV = nRT

or, V = (nR/P) × T.

Thus the slope of the V vs T graph = nR/P where P = 2 atm.

∴ (R/2 atm) = x L mol⁻¹ K⁻¹

or, R = 2x L atm mol⁻¹ K⁻¹

PV = nRT

or, V = (nR/P) × T.

Thus the slope of the V vs T graph = nR/P where P = 2 atm.

∴ (R/2 atm) = x L mol⁻¹ K⁻¹

or, R = 2x L atm mol⁻¹ K⁻¹