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 gives 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 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 
Ideal gas equation its derivation 

Where R is the Universal Gas Constant. This is called ideal gas equation of state for 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} 
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 × 10^{3} cm^{3})/(1 mol × 273 K)
= 8.314 × 10^{7} dyne cm^{2} mol^{1} K^{1} 

Again, Work (W) = Force(F) × Displacement(d),
So, erg = dyne cm^{2}.
Thus, R = 8.314 × 10^{7} erg mol^{1} K^{1} 

We Know That, 1 J = 10^{7} erg,
Thus the vale 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} 
Physical 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

∴ R = (force×length⁻²×length³)/(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 a gas when its temperature is raised by one kelvin.
Determination of Molar mass from 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).
∴ P = dRT/M 
Number of Gas Molecule Present in 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′ k T 

Where N′ = number of molecules present per unit Volume and
k = Boltzmann Constant = R/N₀
= 1.38 × 1016 erg molecule⁻¹ K⁻¹
Ideal Gas Low problems and solutions:
 Problem 1:

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 2:

Derive the value of R when, (a) pressure is expressed in 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 3:

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 4:

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
 Problem 5:

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:
 PV = nRT
or, PV = (g/M)RT
or, M = (g/PV) × RT
= (g × R × T)/(P × V)
= (12.8 × 0.082 × 300)/{(750/760) × 10}
= 31.91
 Problem 6:
 Calculate the pressure excreted on the walls of a 3 litre of flask when 7 gms of nitrogen are introduced into the same 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
 Problem 7:

Calculate the number of molecules present per ml of an ideal gas maintained at pressure of 7.6 × 10^{3} mm of Hg at 0°C.
 Solution:

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}