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*

*Ideal Gas Equation its derivation*

- Boyle's law, V ∝ 1/P When n and T are constant for a gas.

- Charl's Low, V ∝ T When n and P are constant for a gas.

- Avogadro's Low, 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

∴ PV = nRT

or, V = R × (1/P) × T × n

∴ PV = nRT

Ideal gas equation |

- Where R is the Universal gas constant. This is called the

**of state for an**

*ideal gas equation***.**

*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)

= 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

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,

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

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*

*Significance of universal gas constant*

- The universal gas constant 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*

*Molar mass from the ideal gas equation*

The

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).

**is,***ideal gas equation*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 |

###
*Number of gas molecule present in an ideal gas*

*Number of gas molecule present in an ideal gas*

The

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

**for n mole gas is,***ideal gas equation*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

###
*Ideal gas equation problems and solutions*

*Ideal gas equation problems and solutions*

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

^{3}.

*Answer*- 61.54 Torr dm

^{3}mol

^{-1}K

^{-1}

*Problem*- Derive the value of R when, (a) pressure is expressed in the atom, and volume in cm

^{3}and (b) Pressure in dyne m

^{-2}and volume mm

^{3}.

*Answer*- (a) 82.05 atm cm

^{3}mol

^{-1}K

^{-1}

- (b) 8.314 × 10

^{14}dyne m

^{-2}mm

^{3}mol

^{-1}K

^{-1}

*Problem*- Find the Molar mass of ammonia at 5 atm pressure and 30

^{0}C temperature (Density of ammonia = 3.42 gm lit

^{-1}).

*Answer*- 17 gm mol

^{-1}For the details solution see

*Problem*- 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?

*Answer*
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

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

*Problem*- Calculate the number of molecules present per ml of an

**maintained at a pressure of 7.6 × 10**

*ideal gas*^{-3}mm of Hg at 0°C.

*Answer*- We have given that, V = 1ml = 10

^{-6}dm

^{3}

- 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

= 4.46 × 10⁻¹³ mol

- Hence the number of molecules, N = n N

_{0}

- = (4.46 × 10

^{-13}mol)(6.023 × 10

^{23}mol

^{-1}) = 2.68 × 10

^{-11}

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

*Answer*- 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

or, P = {0.1349 × 0.082 × (273+50)}/5

= 0.715 atm

*Problem*- At 2 atm constant pressure slope of a one-mole

**in V vs T graph is x L mol⁻¹ K⁻¹. Find out the value of R by x.**

*ideal gas*

*Answer*
The

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

**for 1-mole gas is,***ideal gas equation*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